# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.11

Exercise 7.11

Direction (Q. 1 to 19) : By using the properties of definite integrals, evaluate the integrals.

$$\textbf{1.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\textbf{cos}^\textbf{2}\textbf{ xdx.}\\\textbf{Sol.\space}\text{Let space I}=\int^{\frac{\pi}{2}}_{0}\text{cos}^{2}\space\text{x dx}\qquad\text{...(i)}\\\Rarr\space\text{I =}\int^{\frac{\pi}{2}}_{0}\text{cos}^{2}\bigg(\frac{\pi}{2}-x\bigg)dx\\\bigg(\because\space \int^{a}_{0}\text{f(x)dx}=\int^{a}_{0}f(a-x)dx\bigg)\\\Rarr\space\text{I}=\int^{\frac{\pi}{2}}_{0}\text{sin}^{2} xdx\qquad\text{...(ii)}$$

On adding equations (i) and (ii), we get

$$\text{2 I}=\int^{\frac{\pi}{2}}_{0}(\text{sin}^2x + \text{cos}^{2}x)dx\\=\int^{\frac{\pi}{2}}_{0} 1 dx\qquad[\because\space sin^{2}x + cos^{2}x=1]\\=[x]^{\frac{\pi}{2}}_{0}=\frac{\pi}{2}=0\Rarr\space 1=\frac{\pi}{4}$$

$$\textbf{2.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\frac{\sqrt{\textbf{sin \space x}}}{\sqrt{\textbf{sin x} + \sqrt{\textbf{cos x}}}}\textbf{dx.}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{\frac{\pi}{2}}_{0}\frac{\sqrt{\textbf{sin x}}}{\sqrt{\textbf{sin x}} + \sqrt{\textbf{cos x}}}dx\qquad\text{...(i)}\\\text{Then,\space}\text{I}=\\\int^{\frac{\pi}{2}}_{0}\frac{\sqrt{\text{sin}\bigg(\frac{\pi}{2}-x\bigg)}}{\sqrt{\text{sin}\bigg(\frac{\pi}{2}+x\bigg)} + \sqrt{\text{cos}\bigg(\frac{\pi}{2}-x\bigg)}}dx\\\bigg(\because\space\int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx\bigg)$$

$$\Rarr\space \text{I = }\int^{\frac{\pi}{2}}_{0}\frac{\sqrt{\text{cos x}}}{\sqrt{\text{cos x} + \sqrt{\text{sin x}}}}dx\qquad\text{...(ii)}\\\begin{bmatrix}\because\space \text{sin}\bigg(\frac{\pi}{2}-x\bigg)=\text{cos x}\\\text{and cos}\bigg(\frac{\pi}{2}-x\bigg)= \text{sin x}\end{bmatrix}$$

On adding equations (i) and (ii), we get

$$\text{2I}=\int^{\frac{\pi}{2}}_{0}\frac{\sqrt{\text{sin x}}+\sqrt{\text{cos x}}}{\sqrt{\text{sin x}}+\sqrt{\text{cos x}}}dx\\=\int^{\frac{\pi}{2}}_{0}\text{1 dx}=[x]^{\frac{\pi}{2}}_{0}=\frac{\pi}{2}-0\\\Rarr\space\text{I}=\frac{\pi}{4}$$

$$\textbf{3.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\frac{\textbf{sin}^{\frac{\textbf{3}}{\textbf{2}}}\textbf{x}}{\textbf{sin}^{\frac{\textbf{3}}{\textbf{2}}}\textbf{x}+\textbf{cos}^{\frac{\textbf{3}}{\textbf{2}}}x}dx.\\\textbf{Sol.}\space \text{Let\space I}=\int^{\frac{\pi}{2}}_{0}\frac{\text{sin}^{\frac{3}{2}}}{\text{sin}^{\frac{3}{2}}x + \text{cos}^\frac{3}{2}x}dx\space\text{...(i)}\\\Rarr\space \text{I}=\\\int^{\frac{\pi}{2}}_{0}\frac{\text{sin}^{\frac{3}{2}}\bigg(\frac{\pi}{2}-x\bigg)}{\text{sin}^{\frac{3}{2}}\bigg(\frac{\pi}{2}-x\bigg)+\text{cos}^{\frac{3}{2}}\bigg(\frac{\pi}{2}-x\bigg)}dx\\\bigg(\because\space\int^{a}_{0}\text{f(x)dx}=\int^{a}_{0}f(a-x)dx\bigg)\\=\int^{\frac{\pi}{2}}_{0}\frac{\text{cos}^{\frac{3}{2}}x}{\text{cos}^\frac{3}{2}x+\text{sin}^\frac{3}{2}x}dx\space\text{...(ii)}$$

$$\begin{bmatrix}\because\space \text{sin}\bigg(\frac{\pi}{2}-x\bigg)=\text{cos x}\\\text{and cos}\bigg(\frac{\pi}{2}-x\bigg)=\text{sin x}\end{bmatrix}$$

On adding equations (i) and (ii), we get

$$\text{2I} = \int^{\frac{\pi}{2}}_{0}\frac{\text{sin}^{\frac{3}{2}x} + \text{cos}^\frac{3}{2} x}{\text{sin}^{\frac{3}{2}}x + \text{cos}^{\frac{3}{2}}x}dx\\=\int^{\frac{\pi}{2}}_{0} 1 dx\\=[x]^{\frac{\pi}{2}}_{0}=\frac{\pi}{2}-0\\\Rarr\space \text{I}=\frac{\pi}{4}$$

$$\textbf{4.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\frac{\textbf{cos}^{\textbf{5}}\textbf{x}}{\textbf{sin}^\textbf{5}\textbf{x}+\textbf{cos}^{\textbf{5}}\textbf{x}}\textbf{dx}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{\frac{\pi}{2}}_{0}\frac{\text{cos}^5x}{\text{sin}^5x + \text{cos}^5 x}dx\qquad\text{...(i)}\\\Rarr\space \text{I = }\\\int^{\frac{\pi}{2}}_{0}\frac{\text{cos}^5\bigg(\frac{\pi}{2}-x\bigg)}{\text{sin}^5\bigg(\frac{\pi}{2}-x\bigg) + \text{cos}^5\bigg(\frac{\pi}{2}-x\bigg)}dx\\\begin{bmatrix}\because\space\int^{a}_{0}\text{f(x)dx}=\int^{a}_{0}f(a-x)dx\end{bmatrix}\\=\int^{\frac{\pi}{2}}_{0}\frac{\text{sin}^5 x}{\text{cos}^5x + \text{sin}^5 x}dx\space\text{...(ii)}$$

On adding equations (i) and (ii), we get

$$\text{2I}=\int^{\frac{\pi}{2}}_{0}\frac{\text{cos}^5x + \text{sin}^{5}x}{\text{cos}^{5}x + \text{sin}^{5}x}dx\\=\int^{\frac{\pi}{2}}_{0} 1dx\\=[x]^{\frac{\pi}{2}}_{0}=\frac{\pi}{2}-0\Rarr \text{I}=\frac{\pi}{4}$$

$$\textbf{5.\space}\int^{\textbf{5}}_{\textbf{\normalsize-5}}\textbf{|x+2|dx}\\\textbf{Sol.}\space\text{Let}\space\text{I = }\int^{5}_{\normalsize-5}|x+2|dx$$

It can be seen that (x + 2) ≤ 0 on [– 5, – 2] and (x + 2)  ≥ 0 on [– 2, 5].

$$\therefore\space\text{I}=\int^{5}_{\normalsize-5}-(x+2)dx+\int^{5}_{\normalsize-2}(x+2)dx\\\bigg[\because\space\int^{b}_{a} f(x)dx=\int^{c}_{a}f(x)dx + \int^{b}_{c}f(x)dx\bigg]\\\Rarr\space \text{I} = -\bigg[\frac{x^2}{2}+2x\bigg]^{\normalsize-2}_{\normalsize-5} + \bigg[\frac{x^2}{2}+2x\bigg]^{5}_{\normalsize-2}$$

$$=-\bigg[\frac{(\normalsize-2)^2}{2}+2(\normalsize-2)-\frac{(\normalsize-5)^2}{2}-2(-5)\bigg]+\\\bigg[\frac{(5)^2}{2}+2(5)-\frac{(-2)^2}{2}-2(-2)\bigg]\\=-\bigg[2-4-\frac{25}{2}+10\bigg]+\bigg[\frac{25}{2}+10-2+4\bigg]\\=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4=29$$

Note : In an absolute integral function, please careful while breaking the limit.

$$\textbf{6.\space}\int^{\textbf{8}}_{\textbf{2}}\space\textbf{|x-5|dx.}\\\textbf{Sol.\space}\text{Let\space} \text{I = }\int^{8}_{2}|x-5|dx$$

It can be seen that (x – 5) ≤ 0 on [2, 5] and (x – 5) ≥ 0 on [5, 8].

$$\because\space\int^{b}_{a}f(x)dx=\int^{c}_{a}f(x)dx+\int^{b}_{c}f(x)dx\\=\therefore\space\text{I = }\int^{5}_{2}\lbrace-(x-5)dx + \int^{8}_{5}(x-5)dx\rbrace\\=\bigg[5x-\frac{x^2}{2}\bigg]^{5}_{2}+\bigg[\frac{x^2}{2}-5x\bigg]^{8}_{5}\\=\bigg[\bigg(25-\frac{25}{2}\bigg)-\bigg(10-\frac{4}{2}\bigg)\bigg]+\\\bigg[\bigg(\frac{64}{2}-40\bigg)-\bigg(\frac{25}{2}-25\bigg)\bigg]\\=\frac{25}{2}-8-8+\frac{25}{2}$$

= 25 – 16 = 9

$$\textbf{7.}\space\int^{\textbf{1}}_{\textbf{0}}\textbf{x(1-x)}^\textbf{n} \textbf{dx.}\\\textbf{Sol.\space}\text{Let}\space \text{I} = \int^{1}_{0}x(1-x)^n dx\\\Rarr\space\text{I}=\int^{1}_{0}(1-x)\lbrace1-(1-x)\rbrace^n dx\\\bigg[\because\space\int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx\bigg]\\=\int^{1}_{0}(1-x)x^n dx=\int^{1}_{0}(x^n-x^{n+1})dx\\=\bigg[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\bigg]^{1}_{0}=\bigg[\frac{1}{n+1}-\frac{1}{n+2}\bigg]-0\\=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}$$

$$\textbf{8.\space}\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}\textbf{log}\textbf{(1+tan x) dx}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{\frac{\pi}{4}}_{0}\text{log}(1 + \text{tan\space x} )dx\space\text{...(i)}\\\Rarr\space\text{I}=\int^{\frac{\pi}{4}}_{0}\text{log}\bigg[1 + \text{tan}\bigg(\frac{\pi}{4}-x\bigg)\bigg]dx\\\bigg[\because\space \int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx\bigg]\\=\int^{\frac{\pi}{4}}_{0}\text{log}\bigg(1+\frac{\text{1 - tanx }}{\text{1 + tan x}}\bigg)dx\\\begin{bmatrix}\because\space\text{tan}(\text{A-B})=\frac{\text{tan A - tan B}}{\text{1 + tan A tan B}},\\\text{here A = }\frac{\pi}{4}, \text{B = x}\end{bmatrix}$$

$$=\int^{\frac{\pi}{4}}_{0}\text{log}\bigg(\frac{2}{\text{1 + tan x}}\bigg)dx\\=\int^{\frac{\pi}{4}}_{0}\lbrace\text{log 2 - log(1+tan x)}\rbrace dx\\\text{...(ii)}\\\bigg[\because\space\text{log}\bigg(\frac{m}{n}\bigg)=\text{log m - log n}\bigg]$$

On adding equations (i) and (ii), we get

$$\text{2I} = \int^{\frac{\pi}{4}}_{0}\text{log 2x}\\=\text{log 2}\int^{\frac{\pi}{4}}_{0} 1 dx = log \space 2[x]^{\frac{\pi}{4}}_{0}\\=\text{(log 2)}\bigg(\frac{\pi}{4}-0\bigg)\\\Rarr\space 2\text{I}=\frac{\pi}{4}\text{log 2}\\\Rarr\space\text{I} = \frac{\pi}{8}\text{log 2}$$

$$\textbf{9.\space}\int^{\textbf{2}}_{\textbf{0}}\textbf{x}\sqrt{\textbf{2-x}}\space\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{2}_{0}x \sqrt{2-x}\space dx\qquad\text{...(i)}\\\text{Also,}\space\text{I}=\int^{2}_{0}(2-x)\sqrt{2-(2-x)}dx\\\bigg[\because\space\int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx\bigg]\\=\int^{2}_{0}(2-x)\sqrt{x}dx=\int^{2}_{0}(2x^{\frac{1}{2}}-x^\frac{3}{2})dx\\=\bigg[\frac{2x^{(\frac{1}{2})+1}}{(\frac{1}{2})+1}-\frac{x^{(\frac{3}{2})+1}}{(\frac{3}{2})+1}\bigg]^{2}_{0}\\=\bigg[\frac{4}{3}x^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}-0\bigg]^{2}_{0}$$

$$=\frac{4}{3}.2^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}-0\\=\frac{4}{3}2\sqrt{2}-\frac{2}{5}4\sqrt{2}\\=\bigg(\frac{8}{3}-\frac{8}{5}\bigg)\sqrt{2}=\bigg(\frac{40-24}{15}\bigg)\sqrt{2}\\=\frac{16\sqrt{2}}{15}$$

$$\textbf{10.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}(\textbf{2 log sin x} - \textbf{log sin 2x})\textbf{dx.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{\frac{\pi}{2}}_{0}(\text{2 log sin x} - \text{log} \space \text{sin 2x})dx\\=\int^{\frac{\pi}{2}}_{0}(\text{log sin}^2x - \text{log sin 2x})dx\\\lbrack\because\space\text{m log n = log n}^m\rbrack\\=\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{sin}^2x}{\text{sin 2x}}\bigg)dx\\\bigg[\because\space \text{log m - log n = log}\frac{m}{n}\bigg]\\=\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{sin}^2x}{\text{2 sin xcos x}}\bigg)dx$$

[∵ sin 2x = 2 sin x cos x]

$$=\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{tan x}}{2}\bigg)dx\\=\int^{\frac{\pi}{2}}_{0}\text{log}(\text{tan x})-\text{log 2}dx\\=\int^{\frac{\pi}{2}}_{0}\text{log (tan x)dx}-\int^{\frac{\pi}{2}}_{0}\text{log 2 dx}\\\Rarr\space \text{I = I}_1-(\text{log 2})[x]^{\frac{\pi}{2}}_{0}\\=\text{I}_1-\bigg(\frac{\pi}{2}-0\bigg)\text{log 2}\\=\text{I}_1-\frac{\pi}{2}\text{log 2}\qquad\text{...(i)}\\\text{where,}\space\text{I}_1=\int^{\frac{\pi}{2}}_{0}\text{log}(\text{tan x})dx\qquad\text{...(ii)}$$

$$\Rarr\space\text{I}_1=\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\text{tan}\bigg(\frac{\pi}{2}-x\bigg)\bigg)dx\\\bigg[\because\space\int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx\bigg]\\\Rarr\space\text{I}_1=\int^{\frac{\pi}{2}}_{0}\text{log (cot x)}dx\qquad\text{...(iii)}$$

On adding equations (ii) and (iii), we get

$$2\text{I}_1=\int^{\frac{\pi}{2}}_{0}(\text{log}(\text{tan x}) + \text{log}(\text{cot x}))dx\\=\int^{\frac{\pi}{2}}_{0}\text{log}(\text{tan x cot x})dx\\\lbrack\because\space\text{log m + log n = log (mn)}\rbrack\\=\int^{\frac{\pi}{2}}_{0}\text{log} 1dx = 0\\\Rarr\space \text{I}_1=0\\\bigg(\because\space\text{tan x}=\frac{1}{\text{cot x}}\bigg)$$

On putting the value of I1 in eq. (i), we get

$$\text{I} = 0 - \frac{\pi}{2}\text{log 2}\\=-\frac{\pi}{2}\text{log 2}$$

$$\textbf{11.\space Let\space}\int^{\frac{\pi}{\textbf{2}}}_{-\frac{\pi}{\textbf{2}}}\textbf{sin}^{\textbf{2}}\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I = }\int^{\frac{\pi}{\text{2}}}_{-\frac{\pi}{2}}\text{sin}^2 xdx$$

Here, f(x) = sin2 x

f(– x) = sin2(– x) = [sin (– x)]2

= (– sin x)2 = sin2 x = f(x)

∴ f(x) is an even function.

$$\text{I} = \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{sin}^2 xdx\\= 2\int^{\frac{\pi}{2}}_{0}\text{sin}^2 dx\\\begin{bmatrix}\because\space\int^{a}_{-a}f(x)dx=2\int^{a}_{0}f(x)dx,\\\text{if f(x) is even}\end{bmatrix}\\= 2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{\text{1 - cos 2x}}{2}\bigg]dx$$

(∵ cos 2x = 1 – 2 sin2x)

$$=\int^{\frac{\pi}{2}}_{0}(1-\text{cos 2x})dx\\=\bigg[x-\frac{\text{sin 2x}}{2}\bigg]^{\frac{\pi}{2}}_{0}\\=\bigg[\frac{\pi}{2}-\frac{\text{sin}\pi}{2}\bigg]-(0-0)\\=\frac{\pi}{2}-0=\frac{\pi}{2}$$

$$\textbf{12.}\space\int^{\pi}_{\textbf{0}}\frac{\textbf{x}}{\textbf{1 + sin x}}\textbf{dx}\\\textbf{Sol.}\space\text{Let}\space\text{I} = \int^{\pi}_{0}\frac{x}{\text{1 + sin x}}dx\qquad\text{...(i)}\\\text{Then,\space I =}\int^{\pi}_{0}\frac{\pi-x}{\text{1 + sin(}\pi-x)}dx\\\bigg[\because\space \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx\bigg]\\\Rarr\space\text{I} = \int^{\pi}_{0}\frac{\pi-x}{\text{1 + sin x}}\qquad\text{...(ii)}\\\lbrack\because\space \text{sin}(\pi-x) = \text{sin \space x}\rbrack$$

On adding equations (i) and (ii), we get

$$\text{2I = }\int^{\pi}_{0}\frac{\pi}{(\text{1+ sin x})}dx\\=\pi\int^{\pi}_{0}\frac{1}{\text{(1 + sin x)}}dx\\=\pi\int^{\pi}_{0}\frac{\text{1 - sin x}}{(\text{1 + sin x})(\text{1 - sin x )}}dx$$

(Multiply numerator and denominator by (1 – sin x))

$$\text{2I}=\pi\int^{\pi}_{0}\frac{\text{1 - sin x}}{\text{1 - sin}^2x}dx\\=\pi\int^{\pi}_{0}\frac{1}{\text{cos}^2x}dx-\pi\int^{\pi}_{0}\frac{\text{sin x}}{\text{cos}^2x}dx$$

[∵ sin2 x + cos2 x = 1]

$$\Rarr\space\text{2I} = \pi\int^{\pi}_{0}\text{sec}^2x dx-\\\pi\int^{\pi}_{0}\text{sec x.tan x}dx$$

$$\Rarr\space \text{2I = }\pi[\text{tan x - sec x}]^{\pi}_{0}$$

$$\Rarr\space\text{2I = }\pi[\text{tan}\space\pi - \text{sec}\space \pi - \\(\text{tan 0 - sec 0})]\\\Rarr\space 2\text{I} = \pi[0+1-0+1]\\\Rarr\space 2\text{I}= 2\pi\Rarr\space \text{I}=\pi$$

$$\textbf{13.\space}\int^{\frac{\pi}{\textbf{2}}}_{-\frac{\pi}{\textbf{2}}}\textbf{sin}^{\textbf{7}}\textbf{x dx}\\\textbf{Sol.\space}\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{sin}^{7}\space\text{x dx}$$

Here, f(x) = sin7 x

f(– x) = sin7(– x)

= [– sin x]7 = – sin7 x

∴ f(– x) = – f(x)

So, f(x) is an odd function.

$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{sin}^7\space x dx=0\\\begin{bmatrix}\because\space\int^{a}_{-a}f(x)dx=0\space\\\text{if f(x) is an odd function}\end{bmatrix}\\\textbf{Note:}\space\int^{a}_{-a}f(x)dx\\ \begin{cases}\because\space\int^{2a}_{0}\text{f(x)dx,\space if f(x) is an even.}\\0,\space \text{if f(x) is an odd.}\end{cases}$$

$$\textbf{14.}\space\int^{\textbf{2}\pi}_{\textbf{0}}\textbf{cos}^\textbf{5} \textbf{x dx}\\\textbf{Sol.\space}\int^{2\pi}_{0}\text{cos}^{5}x dx= 2\int^{\pi}_{0}\text{cos}^5dx\\\because\space\int^{2a}_{0}f(x) = 2\int^{a}_{0}\text{f(x)dx,}$$

where f(2a – x) = f(x)

Hence 2a = 2x

∴ cos5 (2π – x)dx = cos5 x

= 2 × 0 = 0

$$\begin{pmatrix}\because\space\int^{2a}_{0}f(x)=0, \text{if} f(2a-x)=-f(x);\\\text{here 2a = }\pi\therefore\space\text{cos}^5(\pi-x)=-\text{cos}^5 x\end{pmatrix}$$

$$\textbf{15.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\frac{\textbf{sin x - cos x}}{\textbf{1 + sin x cos x}}\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I = }\int^{\frac{\pi}{2}}_{0}\frac{\text{sin x - cos x}}{\text{1 + sinx cos x}}dx\space\text{...(i)}\\\Rarr\space\text{I} =\int^{\frac{\pi}{2}}_{0}\frac{\text{sin}\bigg(\frac{\pi}{2}-x\bigg)-\text{cos}\bigg(\frac{\pi}{2}-x\bigg)}{1 + \text{sin}\bigg(\frac{\pi}{2}-x\bigg)\text{cos}\bigg(\frac{\pi}{2}-x\bigg)}dx\\\bigg[\because\space\int^{a}_{0}\text{f(x)dx} = \int^{a}_{0} f(a-x)dx\bigg]\\=\int^{\frac{\pi}{2}}_{0}\frac{\text{cos x - sin x}}{\text{1 + cos x sin x}}\space\text{...(ii)}$$

$$\begin{bmatrix}\because\space\text{sin}\bigg(\frac{\pi}{2}-x\bigg)=\text{cos x}\\\text{and cos x}\bigg(\frac{\pi}{2}-x\bigg)=\text{sin x}\end{bmatrix}$$

On adding equations (i) and (ii), we get

$$\text{2I = }\int^{\frac{\pi}{2}}_{0}\frac{0}{\text{1 + sin x cos x}}dx=0\\\Rarr\space\text{I =0}$$

$$\textbf{16.}\space\int^{\pi}_{\textbf{0}}\textbf{log}\textbf{(1 + cos x)dx}\\\textbf{Sol.\space}\text{Let}\space \text{I}=\int^{\pi}_{0}\text{log}(1 + \text{cos x})dx\\\text{...(i)}\\\Rarr\space\text{I}=\int^{\pi}_{0}\text{log}\lbrace1 + \text{cos}(\pi-r)\rbrace dx\\\bigg[\because\space\int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx\bigg]\\=\int^{\pi}_{0}\text{log}(1 - cos x)dx\\\lbrack\because\space \text{cos}(\pi-x) = -\text{cos x}\rbrack\qquad\text{...(ii)}\\=\int^{\pi}_{0}\text{log}\begin{Bmatrix}2 \text{sin}^2\bigg(\frac{x}{2}\bigg)\end{Bmatrix}dx$$

$$\bigg[\because\space 1 - cos x = 2 sin^2\frac{x}{2}\bigg]\\=\int^{\pi}_{0}\begin{Bmatrix}\text{log 2 + 2 log}\bigg(\text{sin}\frac{x}{2}\bigg)\end{Bmatrix}dx$$

[∵ log (m n2) = log m + 2 log n]

$$=\int^{\pi}_{0}\text{log 2 dx} + 2\int^{\pi}_{0}\text{log}\bigg(\text{sin}\frac{x}{2}\bigg)dx$$

In the second integral,

$$\text{put}\frac{x}{2}=t\Rarr\space dx=2 dt$$

and limits when x = 0, t = 0, and when x = π, t = π/2

$$\therefore\space\text{I = }\text{log 2}[x]^{\pi}_{0}+2\int^{\frac{\pi}{2}}_{0}\text{log(sin t)2 dt}\\=(log 2)(\pi-0) + 4\bigg(-\frac{\pi}{2}\text{log 2}\bigg)\\\bigg(\because\int^{\frac{\pi}{2}}_{0}\text{log sin x dx}=-\frac{\pi}{2}\text{log 2}\bigg)$$

= – π log 2

$$\textbf{17.\space}\int^{\textbf{a}}_{\textbf{0}}\frac{\sqrt{\textbf{x}}}{\sqrt{\textbf{x}} + \sqrt{\textbf{a-x}}}\textbf{dx.}\\\textbf{Sol.\space}\text{Let}\space\text{I} = \int^{a}_{0}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx\space\text{...(i)}\\\text{I}=\int^{a}_{0}\frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{a - (a-x)}}dx\\\bigg[\because\space\int^{a}_{0}\text{f(x)dx} = \int^{a}_{0}f(a-x)dx\bigg]\\=\int^{a}_{0}\frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}}dx\qquad \text{...(ii)}$$

On adding equations (i) and (ii), we get

$$\Rarr\space \text{2I} = \int^{a}_{0}\frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}}dx,\\=\text{2I}=\int^{a}_{0}\space 1 dx\\=[x]^{a}_{0}=a-0=a\Rarr\space \text{I}=\frac{a}{2}$$

$$\textbf{18.\space}\int^{\textbf{4}}_{\textbf{0}}\textbf{|x-1|dx}\\\textbf{Sol.\space}\text{let}\space\text{I = }\int^{4}_{0}|x-1|dx$$

It can be seen that, (x – 1) ≤ 0 when 0 ≤ x ≤ 1 and (x – 1) ≥ 0 when 1 ≤ x ≤ 4

$$\therefore\space\text{I = }\int^{1}_{0}|x-1|dx + \int^{4}_{1}|x-1|dx\\\bigg[\because\space\int^{b}_{a}\text{f(x)dx} = \int^{c}_{a}\text{f(x)dx} +\int^{b}_{a}f(x)dx\bigg] \\=\int^{1}_{0}(1-x)dx + \int^{4}_{1}(x-1)dx \\=\bigg[x-\frac{x^2}{2}\bigg]^{1}_{0} + \bigg[\frac{x^2}{2}-x\bigg]^{4}_{1}\\=\bigg(1-\frac{1}{2}\bigg)-0 + \bigg(\frac{4^2}{2}-4\bigg)-\\\bigg(\frac{1}{2}-1\bigg)\\=\frac{1}{2} + 4+\frac{1}{2}=5$$

$$\textbf{19.\space\text{show that}}\int^{\textbf{a}}_{\textbf{0}}\textbf{f(x)g(x) dx =}\\ 2\int^{\textbf{a}}_{\textbf{0}}\textbf{f(x)dx,}\space\textbf{if f and g}$$

are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4.

$$\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{a}_{0}\text{f(x)g(x)dx}\space\text{...(i)}\\\Rarr\space\text{I}=\int^{a}_{0}f(a-x)g(a-x)dx\\\bigg[\because\space\int^{a}_{0}\text{f(x)dx}=\int^{a}_{0}f(a-x)dx\bigg]\\\Rarr\space\text{I = }\int^{a}_{0}f(x)\lbrace4-g(x)\rbrace dx\space\text{...(ii)}$$

[∵ f (x) = f(a – x) and g(x) + g(a – x) = 4 (given)]

On adding equations (i) and (ii), we get

$$\text{2I = }\int^{a}_{0} 4\space f(x)dx\\\Rarr\space \text{I = 2}\int^{a}_{0}\text{f(x)dx.}$$

Hence proved.

20. The value of

$$\int^{\frac{\pi}{\textbf{2}}}_{-\frac{\pi}{\textbf{2}}}(\textbf{x}^\textbf{3}\textbf{ + xcos x + tan}^\textbf{5}\textbf{x+1}\textbf{)dx}\space\textbf{is:}$$

(a) 0

(b) 2

(c) p

(d) 1

Sol. (c) π

$$\text{Let\space I}=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(x^3+x cos x + tan^{5}x+1)dx\\\Rarr\space\text{I}=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}x^3dx + dx + \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{x cos x dx}+\\\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{tan}^{5}x dx + \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{1 dx}$$

We know that

$$\int^{a}_{-a}\text{f(x)dx} = \begin{cases} 2\int^{a}_{0}f(x)dx,\space\text{if f(x) is even}\\0,\space \text{if f(x) id odd}\end{cases}\\\therefore\space\text{I =} 0+0+0+2\int^{\frac{\pi}{2}}_{0}1 dx.$$

[∵ x3, x cos x and tan5 (x) are odd functions.]

$$\therefore\space\text{I} = 2[x]^{\frac{\pi}{2}}_{0}=\frac{2\pi}{2}=\pi.$$

$$\textbf{21. The value of}\\\space\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\textbf{log}\bigg(\frac{\textbf{4 + 3 sin x}}{\textbf{4 + 3 cos x}}\bigg)\textbf{dx}\space\textbf{is:}$$

(a) 2

$$\textbf{(b)}\space\frac{3}{4}$$

(c) zero

(d) – 2

Sol. (c) 0

$$\text{Let}\space\text{I} = \int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{4+3 sin x}}{\text{4 + 3\space cos x}}\bigg)dx\space\\\text{...(i)}\\\Rarr\space\text{I} = \int^{\frac{\pi}{2}}_{0}\text{log}\begin{pmatrix}\frac{4 + 3\text{sin}\bigg(\frac{\pi}{2}-x\bigg)}{\text{4 + 3 cos}\bigg(\frac{\pi}{2}-x\bigg)}\end{pmatrix}dx\\\bigg[\because\space\int^{a}_{0}\text{f(x)dx =}\int^{a}_{0}\text{f(a-x)dx}\bigg]\\\Rarr\space\text{I}=\int^{\frac{\pi}{2}}_{0}\text{log}\bigg(\frac{\text{4 + 3 cos x}}{\text{4 + 3 sin x}}\bigg)dx\space\text{...(ii)}\\\begin{bmatrix}\because\space\text{sin}\bigg(\frac{\pi}{2}-x\bigg)= cos x\\\text{and cos}\bigg(\frac{\pi}{2}-x\bigg)=\text{sin x}\end{bmatrix}$$

On adding equations (i) and (ii), we get

$$\text{2I = }\int^{\frac{\pi}{2}}_{0}\begin{bmatrix}\text{log}\bigg(\frac{4 + 3\space sin\space x }{\text{4 + 3 cos x}}\bigg)+\\\text{log}\bigg(\frac{\text{4 + 3 cos x}}{\text{4 + 3 sin x}}\bigg)\end{bmatrix}dx\\\Rarr\space\text{2I} = \int^{\frac{\pi}{2}}_{0}\text{log}\begin{pmatrix}\frac{4 + 3 \space sin x}{4 + 3\space cos x}×\frac{\text{4 + 3 cos x}}{4 + 3\space sin x}\end{pmatrix}dx\\\lbrack\because\space \text{log m + log n = log mn}\rbrack\\\Rarr\space\text{2I} = \int^{\frac{\pi}{2}}_{0}\text{log 1 dx}\\\Rarr\space \text{2I =}\int^{\frac{\pi}{2}}_{0}0 dx\space (\because log\space 1 =0)$$

$$\Rarr\space\text{I = 0.}$$