NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.3
Exercise 7.1 Solutions 22 Questions
Exercise 7.2 Solutions 39 Questions
Exercise 7.3 Solutions 24 Questions
Exercise 7.4 Solutions 25 Questions
Exercise 7.5 Solutions 23 Questions
Exercise 7.6 Solutions 24 Questions
Exercise 7.7 Solutions 11 Questions
Exercise 7.8 Solutions 6 Questions
Exercise 7.9 Solutions 22 Questions
Exercise 7.10 Solutions 10 Questions
Exercise 7.11 Solutions 21 Questions
Miscellaneous Exercise on Chapter 7 Solutions 44 Questions
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 (xex) + 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 (xex) + C
