NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.3

Exercise 7.3

Find the integrals of the functions.

1. sin² (2x + 5)

$$\textbf{Sol.}\int\text{sin}^2(2x+5)dx=\\\int\frac{\text{I - cos 2}(2x+5)}{2}dx\\=\int\frac{\text{1 - cos (4x+10)}}{2}dx\\\bigg(\because\space\text{sin}^2x=\frac{1-\text{cos 2x}}{2}\bigg)\\=\int\frac{1}{2}dx-\frac{1}{2}\int\text{cos}(4x+10)dx\\=\frac{1}{2}x-\frac{1}{2}\frac{\text{sin}(4x+10)}{4}+\text{C}\\\bigg[\because\space\int\text{cos}(ax+b)dx=\frac{\text{sin}(ax+b)}{a}\bigg]$$

$$=\frac{1}{2}x-\frac{\text{sin}(4x+10)}{8}+\text{C}$$

2. sin 3x cos 4x

Sol. ∫ sin 3x cos 4x dx

$$=\frac{1}{2}\int 2\space\text{sin 3x cos 4x}dx\\=\frac{1}{2}\int[\text{sin}(3x+4x) + \text{sin}(3x-4x)]dx\\\lbrack\because\text{2 sin A cos B}=\text{sin}(A+B) + \text{sin}(A-B)\rbrack\\$$

$$=\frac{1}{2}\int(\text{sin 7x - sin x})dx\\\lbrack\because\space \text{sin}(-\theta)=-\text{sin}\theta\rbrack\\=\frac{1}{2}\bigg(\frac{-\text{cos 7x}}{7} + \text{cos x}\bigg)+\text{C}\\\bigg(\because\space \int\text{sin ax dx}=\frac{-\text{cos ax}}{a}\bigg)\\=\frac{-\text{cos} 7x}{14}+\frac{-\text{cos} x}{2}+\text{C}\\\bigg(\because\space\int\text{sin ax dx}=\frac{-\text{cos ax}}{a}\bigg)\\=\frac{-\text{cos}7x}{14}+\frac{\text{cos x}}{2}+\text{C}$$

3. cos 2x cos 4x cos 6x

Sol. ∫ cos 2x cos 4x cos 6x dx

$$=\int\text{cos 2x}\bigg[\frac{1}{2}\lbrace\text{cos}(4x+6x)+\text{cos}(4x - 6x)\rbrace\bigg]dx\\\lbrack\because\space\text{2 cos A cos B}=\text{cos}(A+B) + \text{cos}(A-B)\rbrack\\=\frac{1}{2}\space\int\lbrack\text{cos 2x cos 10x + cos 2x cos(-2x)}\rbrack dx$$

$$=\frac{1}{2}\int\lbrack\text{cos 2x cos 10x + cos}^22x\rbrack dx\\\lbrack\because\space \text{cos}(-\theta)=\text{cos}\space\theta\rbrack\\=\frac{1}{2}\int\bigg[\frac{1}{2}\lbrace\text{cos}(2x+10x) +\text{cos}(2x -10x)\rbrace\bigg]+\\\bigg(\frac{1+\text{cos 4x}}{2}\bigg)dx$$

[∵ cos 2θ = 2 cos² θ – 1]

$$=\frac{1}{4}\int\lbrack\text{cos 12 x + cos 8x + 1 + cos 4x}\rbrack dx\\=\frac{1}{4}\bigg[\frac{\text{sin 12 x}}{12} + \frac{\text{sin 8x}}{8}+x + \frac{\text{sin 4x}}{4}\bigg]+\text{C}$$

4. sin³ (2x + 1)

Sol. ∫ sin³ (2x + 1) dx = ∫ sin² (2x + 1) . sin(2x + 1) dx

= ∫ [1 – cos² (2x + 1)] sin (2x + 1) dx

(∵ sin² x = 1 – cos² x)

Let cos (2x + 1) = t

$$\Rarr\space -2\space\text{sin}(2x+1)dx=dt\\\Rarr\space \text{sin}(2x+1)dx=-\frac{dt}{2}\\\therefore\space\int\text{sin}^3(2x+1)dx=\frac{-1}{2}\int(1-t^2)dt\\=\frac{-1}{2}\bigg( t-\frac{t^3}{3}\bigg)+\text{C}\\=\frac{-1}{2}\bigg(\text{cos}(2x+1)-\frac{\text{cos}^3(2x+1)}{3}\bigg)+\text{C}\\=\frac{\text{- cos}(2x+1)}{2} + \frac{\text{cos}^3(2x+1)}{6}+\text{C} $$

5. sin³ x . cos³ x

Sol. Let I = ∫ sin³ x . cos³ x dx

Let cos x = t

$$\Rarr\space\text{- sin x}=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{- sin x}}\\\therefore\space\text{I}=\int\text{sin}^3 x.t^3\frac{dt}{\text{- sin x}}$$

= – ∫ sin² x.t³dt

= – ∫ t³ (1 – cos²) dt

= – ∫ t³ .(1 – t²) dt

$$=-\int[t^3-t^5]dt=-\bigg[\frac{t^4}{4}-\frac{t^6}{6}\bigg]+\text{C}\\=\frac{\text{cos}^6x}{6}-\frac{\text{cos}^4x}{4}+\text{C}$$

6. sin x sin 2x sin 3x

Sol. ∫ sin x sin 2x sin 3x dx

$$=\frac{1}{2}\int(2\space\text{sin 3x sin x})\space\text{sin 2x}\space dx\\\begin{bmatrix}\because\space \text{2 sin A sin B}=\\\text{cos}(A-B)-\text{cos}(A+B)\end{bmatrix}$$

$$=\frac{1}{2}\int\lbrace\text{cos 2x - cos 4x}\rbrace\space\text{sin 2x}dx\\=\frac{1}{4}\int\lbrace 2\space\text{sin 2x cos 2x - 2 cos 4x sin 2x}\rbrace dx\\=\frac{1}{4}\int\lbrace(\text{sin 4x + sin 0})-(\text{sin 6x - sin 2x})\rbrace dx$$

$$\begin{bmatrix}\because\space\text{2 sin A cos B} = \text{sin}(A+B) + \text{sin}(A-B)\\\text{2 cos A sin B} = \text{sin}(A+B)-\text{sin}(A-B)\end{bmatrix}$$

$$=\frac{1}{4}\int(\text{sin 4x - sin 6x + sin 2x})dx\\=\frac{1}{4}\begin{Bmatrix}\frac{\text{- cos 4x}}{4} - \frac{(\text{- cos 6x})}{6} + \frac{(-\text{cos 2x})}{2}\end{Bmatrix}+\text{C}\\\bigg(\because\space\int\text{sin axdx}=-\frac{\text{cos ax}}{a}\bigg)\\=\frac{-\text{cos 4x}}{16}+\frac{\text{cos 6x}}{24}-\frac{\text{cos 2x}}{8}+\text{C}$$

7. sin 4x sin 8x

Sol. ∫ sin 4x sin 8x dx

$$=\frac{1}{2}\int\text{2 sin 8x sin 4x}dx\\=\frac{1}{2}\int(\text{cos 4x - cos 12x})dx\\\begin{bmatrix}\because\space \text{2 sin A sin B}=\\\text{cos}(A-B)-\text{cos}(A+B)\end{bmatrix}\\=\frac{1}{2} \begin{Bmatrix}\frac{\text{sin 4x}}{4}-\frac{\text{sin 12 x}}{12}\end{Bmatrix}+\text{C}$$

$$\textbf{8.}\space\frac{(\textbf{1 + cos x})}{\textbf{(1+ cos x)}}\\\textbf{Sol}.\int\frac{(\text{1- cos x})}{(\text{1 + cos x )}}dx=\\\int\frac{2\text{sin}^2\frac{x}{2}}{\text{2 cos}^2\frac{x}{2}}dx=\\\int\text{tan}^2\frac{x}{2}dx\\\begin{bmatrix}\because\space 1- cos x=2 \text{sin}^2\frac{x}{2}\\\text{and}\space\text{1 + cos x}= 2cos^2\frac{x}{2}\end{bmatrix}$$

$$=\int\bigg(\text{sec}^2\frac{x}{2}-1\bigg)dx\\\lbrack\because \space\text{tan}^2\theta=\text{sec}^2\theta-1\rbrack\\=\int\text{sec}^2\frac{x}{2}dx-\int 1dx=\\\frac{\text{tan}\frac{x}{2}}{\frac{1}{2}}-x+C\\\bigg(\because\space\int\text{sec}^2 axdx=\frac{\text{tan ax}}{a}\bigg)\\=\text{2 tan}\frac{x}{2}+\text{C}$$

$$\textbf{9.\space }\frac{\textbf{cos x}}{\textbf{1 + cos x}}\\\textbf{Sol.}\space\int\frac{\text{cos x}}{\text{1+ cos x}}dx=\\\int\frac{1-1 + cos x}{1 + cos x}dx\\\lbrack\text{add and subtract 1 in numerator}\rbrack\\=\int\frac{1 + cos x}{1 + cos x}dx-\int\frac{1}{\text{1+ cos x}}dx$$

$$=\int 1 dx-\int \frac{1}{\text{2 cos}^2\frac{x}{2}}dx\\=\int 1 dx-\frac{1}{2}\int\text{sec}^2\frac{x}{2}dx\\=x-\frac{1}{2}.2\text{tan}\frac{x}{2}+\text{C}\\= x-\text{tan}\frac{x}{2}+\text{C}\\\bigg(\because\space\int\text{sec}^2 axdx=\frac{\text{tan ax}}{a}\bigg)$$

$$\textbf{10.\space sin}^\textbf{4} \textbf{x}\\\textbf{Sol.}\space\int\text{sin}^4 xdx = \int(\text{sin}^2x)^2dx\\=\int\bigg(\frac{1-\text{cos 2x}}{2}\bigg)^2dx\\\bigg(\because\space \text{sin}^2x=\frac{1-\text{cos 2x}}{2}\bigg)\\=\frac{1}{4}\int(1- cos 2x)^2 dx\\=\frac{1}{4}\int 1+\text{cos}^22x-2 cos 2x dx\\=\frac{1}{4}\begin{bmatrix}\int 1 dx + \int\frac{\text{1+ cos 4x}}{2}dx\\-2\int\text{cos 2x dx}\end{bmatrix}$$

$$\bigg(\because\space\text{cos}^2 x=\frac{\text{1 + cos 2x}}{2}\bigg)\\=\frac{1}{4}\begin{bmatrix}\int 1dx + \frac{1}{2}\lbrace\int 1 dx + \int\text{cos 4x\space dx}\\-2\int\text{cos 2x dx}\rbrace\end{bmatrix}$$

$$=\frac{1}{4}\bigg[ x + \frac{1}{2}\begin{Bmatrix}x+\frac{\text{sin 4x}}{4}\end{Bmatrix}-\frac{2 sin 2x}{2}\bigg]+\text{C}\\=\frac{1}{4}x+\frac{x}{8}+\frac{\text{sin 4x}}{32}-\frac{\text{sin 2x}}{4} + \text{C}\\=\bigg(\frac{2x+x}{8}\bigg) + \frac{\text{sin 4x}}{32}-\frac{\text{sin 2x}}{4}+ \text{C}\\=\bigg[\frac{3x}{8}+\frac{\text{sin 4x}}{32}-\frac{\text{sin 2x}}{4}\bigg]+\text{C}$$

11. cos⁴ 2x

Sol. ∫ cos⁴ 2x dx = ∫ (cos² 2x)² dx =

$$\int\bigg(\frac{\text{1 + cos 4x}}{2}\bigg)^2dx\\=\frac{1}{4}\int(1 + cos 4x)^2 dx\\=\frac{1}{4}\int(1+ \text{cos}^2 4x + 2 cos 4x)dx\\=\frac{1}{4}\lbrack\int 1dx + \int \text{cos}^2 4x dx + 2\int\text{cos 4x}\rbrack\\=\frac{1}{4}\bigg[\int 1 dx + \int\frac{1+cos 8x}{2}dx+2\int\text{cos 4x dx}\bigg]$$

$$\bigg(\because \text{cos}^2x=\frac{\text{1+ cos 2x}}{2}\bigg)\\=\frac{1}{4}\begin{bmatrix}\int 1 dx + \frac{1}{2}\int 1 dx +\\\int\text{cos 8x dx} + 2\int\text{cos 4x dx}\end{bmatrix}\\=\frac{1}{4}\bigg[x+\frac{1}{2}\begin{Bmatrix} x+\frac{\text{sin 8x}}{8}\end{Bmatrix}+\frac{\text{2 sin 4x}}{4}\bigg]+\text{C}\\=\frac{1}{4}x+\frac{x}{8}+\frac{\text{sin 8x}}{64}+\frac{\text{sin 4x}}{8}+\text{C}\\\frac{1}{4}x+\frac{x}{8}+\frac{\text{sin 8x}}{64}+\frac{\text{sin 4x}}{8}+\text{C}\\=\bigg[\frac{3x}{8}+\frac{\text{sin 8x}}{64} + \frac{\text{sin 4x}}{8}\bigg]+\text{C}$$

$$\textbf{12.}\space\frac{\textbf{sin}^\textbf{2}\textbf{x}}{\textbf{(1 + cos x)}}\\\textbf{Sol.}\space \int\frac{\text{sin}^2x}{(\text{1 + cos x})}dx\\=\int\frac{(1-\text{cos}^2x)}{\text{(1+ cosx)}}dx$$

(∵ sin² x = 1 – cos² x)

$$=\int\frac{(1 + cos x)(1 - cos x)}{(1 + cos x)}dx\\\int\space (1 - cos x)dx\\=\int 1 dx - \int\text{cos x dx}$$

= x – sin x + C

$$\textbf{13.}\space \frac{\textbf{cos 2x} -\textbf{cos 2}\alpha}{\textbf{cos x - cos}\alpha}\\\textbf{Sol.}\space\int\frac{\text{cos 2x - cos 2}\alpha}{\text{cos} x -\text{cos}\alpha}dx\\=\int\frac{(2\space\text{cos}^2x-1)-(2 cos^2\alpha -1)}{(\text{cos x - cos}\alpha)}dx\\(\because\space \text{cos 2x = 2 cos}^2x-1)\\=\int\frac{2 cos ^2 x-1-2 \text{cos}^2\alpha+1}{(\text{cos x - cos}\alpha)}dx\\\int\frac{2(cos^2x - cos ^2\alpha)}{(\text{cos x - cos}\alpha)}dx\\=2\int\frac{(cos x - cos \alpha)(cos x + cos\alpha)}{(cos x - cos \alpha)}dx$$

[∵ a² – b² = (a + b)(a – b)]

= 2[∫ cos x dx + cos α ∫ 1 dx]

= 2[sin x + cos α.x] + C

= 2 sin x + 2x cos α + C.

$$\textbf{14.}\space\frac{\textbf{cos x - sin x}}{\textbf{1 + sin 2x}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{\text{cos x - sin x}}{1 + sin 2x}dx\\=\int\frac{\text{cos x - sin x}}{\text{sin}^2x +\text{cos}^2x + 2 sin xcos x}dx\\\begin{bmatrix}\because\space\text{sin}^2x + \text{cos}^2x=1;\\\text{sin 2x = 2 sinx cos x}\end{bmatrix}\\=\int\frac{\text{cos x - sin x}}{(\text{sin x + cos x})^2}dx\\\text{Let cos x + sin x = t}\\\Rarr\space-\text{sin x + cos x}=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{(cos x - sin x)}}$$

$$\therefore\space\text{I}=\int\frac{\text{cos x - sin x}}{t^2}.\frac{dt}{(\text{cos x - sin x})}\\=\int\frac{1}{t^2}dt=\int t^{-2}dt=\frac{t^{-2+1}}{-2+1}+\text{C}\\=\frac{-1}{\text{cos x + sinx }}+\text{C}$$

15. tan³ 2x sec 2x

Sol. Let I = ∫ tan³ 2x sec 2x dx

$$\text{Let}\space\text{sec 2x = t}\\\Rarr\space\text{2 sec 2x tan 2x}=\frac{dt}{dx}\\\Rarr\space dx = \frac{dt}{\text{2 sec x. tan 2x}}\\\therefore\space \text{I}=\int\text{tan}^3 2x\space\text{sec 2x}\frac{dt}{\text{2 sec 2x . tan 2x}}\\=\frac{1}{2}\int\text{tan}^2 2xdt=\frac{1}{2}\int[\text{sec}^22x-1]dt\\(\because\space\text{tan}^2x = \text{sec}^2x-1)\\=\frac{1}{2}\int\lbrack(t^2-1)dt\rbrack\\=\frac{1}{2}\bigg[\frac{t^3}{3}-t\bigg]+\text{C}$$

$$=\frac{1}{2}\bigg[\frac{\text{sec}^32x}{3}-\text{sec 2x}\bigg]+\text{C}\\=\frac{1}{6}\text{sec}^3\text{2x}-\frac{1}{2}\text{sec 2x}+\text{C}$$

16. tan⁴ x

Sol. Let I = ∫ tan⁴ x dx = ∫ (tan² x)² dx

$$\Rarr\space\text{I}=\int(tan^2x)(tan^2x)dx\\=\int (\text{sec}^2x-1)(tan^2x)dx\\\int\text{sec}^2x\text{tan}^2x dx - \int\text{tan}^2x dx\\=\int\text{sec}^2x\text{tan}^2xdx -\int[\text{sec}^2x-1]dx\\=\int\text{sec}^2x\text{tan}^2x dx-\lbrack\int\text{sec}^2xdx-\int 1 dx\rbrack$$

Now, let I₁ = ∫ sec² x tan² x dx

and I₂ = ∫ sec² x dx – 1 ∫ dx

Then, I = I₁ – I₂ ...(i)

Putting tan x = t

$$\Rarr\space \text{sec}^2x=\frac{dt}{dx}\\\Rarr\space dx = \frac{dt}{\text{sec}^2x}\\\therefore\space\text{I}_1=\int\text{sec}^2x t^2.\frac{dt}{\text{sec}^2x}\\\Rarr\space\text{I}_1=\int t^2 dt=\frac{t^3}{3}+\text{C}_1\\=\frac{\text{tan}^3 x}{3}+\text{C}_1$$

I₂ = ∫ sec² x dx – ∫ 1 dx = tan x – x + C₂

∴ Putting the values of I₁ and I₂ in Eq. (i), we get

$$\text{I}=\frac{\text{tan}^3x}{3}+\text{C}_1-(\text{tan x-x})-\text{C}_2\\\Rarr\space\text{I}=\frac{\text{tan}^3x}{3}-\text{tan x} + x+\text{C}\\\begin{pmatrix}\because\space \text{constant – constant = constant}\\\therefore\space \text{C}_1-\text{C}_1=\text{C}\end{pmatrix}$$

$$\textbf{17.\space}\frac{\textbf{sin}^\textbf{3} \textbf{x} + \textbf{cos}^\textbf{3} \textbf{x}}{\textbf{sin}^\textbf{2}\textbf{x} + \textbf{cos}^\textbf{2} \textbf{x}}\\\textbf{Sol.}\space\frac{\text{sin}^3x + \text{cos}^3 x}{\text{sin}^2x\text{cos}^2x}dx\\=\int\frac{\text{sin}^3x}{\text{sin}^2x\text{cos}^2x}dx+\int\frac{\text{cos}^3x}{\text{sin}^2x\text{cos}^2x}dx\\=\int\frac{\text{sinx}}{\text{cos x cos x}}dx+\int\frac{\text{cos x}}{\text{sin x sin x}}dx$$

= ∫ tan x.sec x dx + ∫ cot x.cosec x dx

= sec x – cosec x + C

$$\textbf{18.}\space \frac{\textbf{cos 2x + 2 sin}^\textbf{2} \textbf{x}}{\textbf{cos}^\textbf{2}\textbf{x}}\\\textbf{Sol.}\space\int\frac{\text{cos 2x + 2 sin}^2 x}{\text{cos}^2x}dx\\=\int\frac{1-2\text{sin}^2x + 2 sin^2 x}{\text{cos}^2x}dx\\(\because \space\text{cos 2x=1-2 sin}^2x)\\=\int\frac{1}{\text{cos}^2x}dx=\int\text{sec}^2x dx$$

= tan x + C

$$\textbf{19.}\space\frac{1}{\textbf{sin x cos}^3\textbf{x}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{\text{sinx cos}^3x}dx\\=\int\frac{\text{cos x}}{\text{sin x}}\text{sec}^4 xdx\\=\int\frac{\text{sec}^2x\text{sec}^2x}{\text{tan x }}dx\\=\int\frac{(1+tan^2x)\text{sec}^2x}{\text{tan x}}dx\\(\because\space \text{sec}^2x=1+\text{tan}^2x)\\\text{Putting tan x = t}\\\Rarr\space\text{sec}^2x=\frac{dt}{dx}$$

$$\Rarr\space dx=\frac{dt}{\text{sec}^2x}\\\therefore\space\text{I}=\int\bigg[\frac{1+ t^2}{t}\bigg]\text{sec}^2x\frac{dt}{\text{sec}^2x}\\=\int\frac{1+t^2}{t}dt=\int\bigg[\frac{1}{t}+t\bigg]dt\\=\text{log}|t| +\frac{t^2}{2}+\text{C}\\=\text{log}|tan x| + \frac{\text{tan}^2x}{2}+\text{C}$$

$$\textbf{20.}\space\int\frac{\textbf{cos 2x}}{\textbf{(cos x + sin x)}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{\text{cos 2x}}{(\textbf{cos x + sinx })^2}\\\textbf{Sol.}\space\text{Let}\space I=\int\frac{\text{cos 2x}}{(\text cos x + sin x)^2}dx\\=\int\frac{\text{cos}^2 x -\text{sin}^2 x}{(\text{cos x + sinx })}dx\\\lbrack\because\space\text{cos 2x = cos}^2 x - sin^{2}x\rbrack\\=\int\frac{(\text{cos x - sin x})(cos x + sin x)}{(\text{cos x + sin x})^2}dx\\=\int\frac{\text{cos x - sinx }}{\text{cos x + sin x}}dx$$

[∵ (a² – b²) = (a – b) (a + b)]

Putting cos x + sin x = t

$$\Rarr\space \text{- sinx + cos x}=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{cos x - sinx }}\\\therefore\space \text{I}=\int\frac{\text{cos x - sin x}}{t}.\frac{dt}{\text{cos x - sinx }}\\=\int\frac{1}{t}dt=\text{log}|t|+\text{C}$$

= log|cos x + sin x| + C.

21. sin^–1 (cos x)

Sol. ∫ sin^–1 (cos x) dx

$$\int\begin{Bmatrix}\frac{\pi}{2}-\text{cos}^{\normalsize-1}(cos \space x)\end{Bmatrix}dx\\\bigg(\because \text{sin}^{\normalsize-1}t + \text{cos}^{\normalsize-1}t=\frac{\pi}{2}\text{for}|t|\leq 1\bigg)\\=\int\bigg(\frac{\pi}{2}-x\bigg)dx=\frac{\pi}{2}x-\frac{x^2}{2}+\text{C}$$

$$\textbf{22.}\space\frac{\textbf{1}}{\textbf{cos(x-a)}\textbf{cos(x-b)}}\\\textbf{Sol.}\space\int\frac{1}{\text{cos}(x-a)\text{cos}(x-b)}dx\\=\frac{1}{\text{sin}(b-a)}\int\frac{\text{sin}(x-a) -(x-b)}{\text{cos}(x-a)\text{cos}(x-b)}dx$$

Multiply by sin (b – a) in numerator and deno-minator both and sin (b – a) = sin[(x – a) – (x – b)].

sin(x-a)cos(x-b)

$$=\frac{1}{\text{sin}(b-a)}\int\frac{-\text{cos}(x-a)\text{sin}(x-b)}{\text{cos}(x-a)\text{cos}(x-b)}dx\\\lbrack\because\space \text{sin}(C-D)= \text{sin C cos D - cos C sin D}\rbrack\\=\frac{1}{\text{sin}(b-a)}\int\begin{bmatrix}\frac{\text{sin}(x-a)\text{cos}(x-b)}{\text{cos}(x-a)\text{cos}(x-b)}-\\\frac{\text{cos}(x-a)\text{sin}(x-b)}{\text{cos}(x-a)\text{cos}(x-b)}\end{bmatrix}dx\\=\frac{1}{\text{sin}(b-a)}\int\lbrace\text{tan}(x-a)-\text{tan}(x-b)\rbrace dx\\=\frac{1}{\text{sin}(b-a)}.\lbrace-\text{log}|\text{cos}(x-a)| + \\\text{log}|\text{cos}(x-b)|\rbrace+\text{C}\\\lbrack\because\space\int\text{tan x dx}=-\text{log}|cos x|\rbrack$$

$$=\frac{1}{\text{sin}(b-a)}\text{log}\begin{vmatrix}\frac{\text{cos}(x-b)}{\text{cos}(x-a)}\end{vmatrix}+\text{C}$$

Choose the correct answer.

$$\textbf{23.}\space\int\frac{\textbf{sin}^\textbf{2}\textbf{x} -\textbf{cos}^\textbf{2}\textbf{x}}{\textbf{sin}^\textbf{2}\textbf{x}\textbf{cos}^\textbf{2}\textbf{x}}\textbf{dx}\\\space\textbf{is equal to :}$$

(a) tan x + cot x + C

(b) tan x + cosec x + C

(c) – tan x + cot x + C

(d) tan x + sec x + C

Sol. (a) tan x + cot x + C

$$\int\frac{\text{sin}^2x - \text{cos}^2x}{\text{sin}^2x\text{cos}^2x}dx=$$

$$\int\bigg(\frac{1}{\text{cos}^2x}-\frac{1}{\text{sin}^2x}\bigg)dx\\=\int(\text{sec}^2x - \text{cosec}^2x)dx\\=\text{tan x + cot x}+\text{C}$$

$$\textbf{24.}\space\int\frac{\textbf{e}^\textbf{x}\textbf{(1+x)}}{\textbf{cos}^\textbf{2}(\textbf{e}^\textbf{x} \textbf{x})}\textbf{dx}\\\textbf{is equal to:}$$

(a) – cot (ex x) + C

(b) tan (xex) + C

(c) tan (ex) + C

(d) cot (ex) + C

Sol. (b) tan (xe^x) + C

$$\text{Let}\space \text{I}=\int\frac{\text{e}^\text{x}\text{(1+x)}}{\text{cos}^2(e^x x)}dx\\\text{Let}\space xe^x=t\\\Rarr(xe^x + e^x)=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{e^x(x+1)}\\\therefore\space\text{I}=\int\frac{e^x(1+x)}{\text{cos}^2t}×\frac{dt}{e^x(1+x)}\\=\int\frac{1}{\text{cos}^2t}dt=\int\text{sec}^2tdt$$

= tan t + C

= tan (xe^x) + C

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