NCERT Solutions for Class 12 Maths Chapter 7 - Integrals
$$\textbf{Ans. }\int sin^2(2x + 5)\space dx. \\ = \int \frac{1-cos(4x+10)}{2} dx \space \space \left[\because sin^2x= \frac{1-cos\space 2x}{2}\right] \\ = \frac{1}{2} \int 1\space dx-\frac{1}{2}\int cos\space (4x + 10)\space dx \\ = \frac{x}{2}-\frac{sin(4x+10)}{8}+C \\ $$
$$\textbf{Ans. } \text{Let}\space \space I = \frac{1}{2}\int(2\space cos\space 2x \space cos\space 4x)\space cos\space 6x \space dx \\ \Rightarrow \space \space I = \frac{1}{2}\int(cos\space 6x + cos2x)cos\space 6x \space dx\\ \left[\because 2\space cos\space A \space cos\space B = cos\space(A+B) + cos(A-B)\right]\\ = \frac{1}{2}\int cos\space 6x \space cos\space 6x\space dx + \frac{1}{2}\int cos\space 2x\space cos\space 6x \space dx \\ = \frac{1}{4}\int2\space cos\space 6x\space cos\space 6x \space dx + \frac{1}{4}\int 2cos\space 2x\space cos\space 6x\space dx\\ = \frac{1}{4}\int (cos\space 12x + cos\space 0) \space dx + \frac{1}{4} \int cos \space 8x + cos\space 4x \space dx \\ = \frac{1}{4}\left[\int(cos \space 12x + 1 + cos 8x + cos4x)dx\right] \\ = \frac{1}{4}\left[\frac{sin\space 12x}{12} + \frac{sin\space 8x}{8} + \frac{sin\space 4x}{4}+ x\right] + C$$
$$\textbf{Ans. } \text{Let }\space \space I = \int \frac{1}{cos(x-a)cos(x-b)}dx \\ \text{Divide and multiply by sin (b – a)} \\ I = \frac{1}{sin(b-a)}\int\frac{sin(b-a)}{cos(x-a)cos(x-b)}dx \\ = \frac{1}{sin(b-a)} \int \frac{sin[(x-a)-(x-b)]}{cos(x-a)cos(x-b)}dx \\ \text{(By adding and subtracting x)} \\ = \frac{1}{sin(b-a)} \int\frac{sin(x-a)cos(x-b)-cos(x-a)sin(x-b)}{cos(x-a)cos(x-b)}dx\\ = \frac{1}{sin(b-a)}\int \frac{sin(x-a)cos(x-b)}{cos(x-a)cos(x-b)}dx-\frac{1}{sin(b-a)}\int \frac{cos(x-a)sin(x-b)}{cos(x-a)cos(x-b)}dx \\= \frac{1}{sin(b-a)}\int tan(x-a)dx – \frac{1}{sin(b-a)}\int tan(x-b)dx \\ = \frac{1}{sin(b-a)} \{-log\space |cos(x-a)|+log|cos(x-b)|\} \left[\because \int tan\space x dx = -log|cos\space x| + C\right] \\ I = \frac{1}{sin(b-a)}log\begin{vmatrix}\frac{cos(x-b)}{cos(x-a)} \end{vmatrix} + C.$$
$$\textbf{Ans. } \text{Given,}\space \int\frac{1}{sec\space x+cosec\space x} dx \\ = \int \frac{1}{\frac{1}{cos\space}+\frac{1}{sin\space x}}dx \\ = \int \frac{1}{\frac{sin\space x +cos\space x}{sin\space x \space cos\space x}}dx \\ = \int \frac{sin\space x \space cos\space x}{sin\space x + cos\space x} dx \\ = \frac{1}{2} \int \frac{2sin\space x \space cos\space x}{sin\space x + cos\space x} dx \space \left[\because \text{Divide and multiple by}\space 2\right] \\ = \frac{1}{2}\int \frac{sin\space 2x}{sin\space x + cos\space x}dx\space [\because \space sin \space 2A \space = 2\space sin\space A \space cos\space A] \\ = \frac{1}{2}\int \frac{(1+ sin\space 2x)-1}{sin\space x + cos\space x}dx \space \text{[add \space and\space subtract \space 1]} \\ =\frac{1}{2} \int \frac{(sin^2x + cos^2x + 2sin\space x\space cos\space x) – 1 }{sin\space x + cos\space x} dx \space \space [\because \space sin^2\space x + cos^2\space x = 1] \\ =\frac{1}{2} \int \frac{(sin\space x + cos\space x)^2 – 1}{sin\space x + cos\space x} dx \\ =\frac{1}{2}\int \frac{(sin\space x + cos\space x)^2}{sin\space x + cos\space x}dx\space -\frac{1}{2}\int \frac{1}{sin\space x + cos\space x}dx \\ = \frac{1}{2}\int (sin\space x + cos\space x)dx – \frac{1}{2} \int \frac{1}{\sqrt{2}\left(\frac{1}{\sqrt{2}}sin\space x + \frac{1}{\sqrt{2}}cos\space x\right)}dx \\ \text{Multiple \& divide by}\space \sqrt{2} \\ = \space \frac{1}{2} \int (sin\space x + cos\space x) \space dx – \frac{1}{2\sqrt{2}}\int \frac{dx}{sin\space x\space cos\frac{\pi}{4} + cos\space x \space sin\space \frac{\pi}{4} } \\ = \frac{1}{2} [-cos\space x + sin\space x]- \frac{1}{2\sqrt{2}}\int cosec\left(x+\frac{\pi}{4}\right) \\ = \frac{-1}{2}cos\space x + \frac{1}{2}sin\space x-\frac{1}{2\sqrt{2}}log\begin{vmatrix}cosec\left(x + \frac{\pi}{4}\right)-cot\left(x+ \frac{\pi}{4}\right) \end{vmatrix} + C$$
$$\textbf{Ans. } \text{Let }\space \space I \space = \space \int \frac{e^{2x}-e^{-2x}}{e^{2x}+ e^{-2x}}dx \\ \text{Multiply and divide by 2} \\ I = \frac{1}{2}\int \frac{2e^{2x}-2e^{-2x}}{e^{2x}+e^{-2x}}dx \\ = \frac{1}{2} log |e^{2x}+e^{-2x}| + C \left[\because \int \frac{f'(x)}{f(x)}dx = log|f(x)| + C\right]$$
$$\textbf{Ans. } \text{Let }\space\space I = \space\int \frac{x^3\space sin(tan^{-1}x^4)}{1+x^8}dx \\ Put \space \space \space tan^{-1}x^4 = t \\ \text{Differentiate w.r.t. x }\\ \Rightarrow \space \space \frac{1}{1+x^8}.4x^3 = \frac{dt}{dx} \\ \frac{x^3}{1+x^8}dx = \frac{dt}{4} \\ I = \frac{1}{4}\int sin\space t\space dt \\ = \frac{1}{4}(-cos\space t) + C \\ \Rightarrow \space \space I = -\frac{1}{4}cos(tan^{-1}x^4) + C$$
$$\textbf{Ans. } \text{Let} \space\space I = \int tan^2 \space 2x\space tan\space 2x\space sec\space 2x\space dx \\=\space \int (sec^2 2x-1)tan\space 2x sec\space 2x\space dx \\ Put \space \space \space sec\space 2x = \space t \\ \text{Differentiate both sides w.r.t. x} \\ 2\space sec\space 2x \space tan\space 2x = \frac{dt}{dx} \\ \therefore \space \space sec\space 2x \space tan\space 2x\space dx= \frac{dt}{2} \\ I = \frac{1}{2}\int(t^2 -1) dt \\ \Rightarrow \space \space \space I = \frac{1}{2}\begin{bmatrix}\frac{t^3}{3}-t\end{bmatrix} + C \\ \Rightarrow \space \space I = \space \frac{1}{2}\begin{bmatrix}\frac{sec^3 2x}{3}-sec\space 2x \end{bmatrix} + C \\ \Rightarrow \space \space I = \frac{sec^3\space 2x}{6}-\frac{sec\space 2x}{2} + C$$
$$\textbf{Ans. } \text{We have,} \space \int \frac{2x}{(x^2+1)(x^2+3)} dx \\ Put \space \space \space x^2 = t \\ \text{Differentiate both side w.r.t. x} \\ 2x \space dx \space = dt \\ \therefore \space \space \space I = \int \frac{dt}{(t+1)(t+3)} \\ \frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3} \\ \text{Multiplication both side by (t + 1) (t + 3)} \\ \text{I = A(t + 3) + B(t +1) \space \space …(1)} \\ \text{Put\space \space t = –1 in equation (1)} \\ \text{then\space\space \space 1 = A(–1 + 3)}\\ \Rightarrow \space \space \space 1 = 2A \\ \Rightarrow \space \space A = \frac{1}{2} \\ \text{If\space \space \space t = – 3} \\ \text{Put the value in equation (1)} \\ \text{then\space \space \space 1 = B(– 3 + 1)} \\ 1 = -2B \\ B = -\frac{1}{2} \\ \therefore \space \space I = \int \frac{\frac{1}{2}}{t+1}dt + \int \frac{-\frac{1}{2}}{t+3}dt\\ \Rightarrow \space \space \space I = \frac{1}{2}\space log | t+1 | – \frac{1}{2}\space log \space |t+3|+ C \\ = \frac{1}{2}\begin{bmatrix}log \begin{vmatrix}\frac{t+1}{t+3} \end{vmatrix} \end{bmatrix}+c \\ \Rightarrow \space \space \space I = \frac{1}{2}log\begin{vmatrix}\frac{x^2 + 1}{x^2 + 3} \end{vmatrix} + C$$
$$\textbf{Ans. } \int \frac{3x-1}{(x-1)(x-2)(x-3)}\space =\space \frac{A}{x-1} + \frac{B}{x-2}+\frac{C}{x-3} \\ \text{3x – 1 = A(x – 2) (x – 3) + B(x – 1) (x – 3)+ C(x –1) (x –2)} \\ \text{If\space \space\space x = 1, then} \\ \text{2 = A(– 1) (– 2)} \\ \Rightarrow \space \space \space \text{2A = 2} \\ \Rightarrow \space \space \text{A = 1} \\ If \space x = 2,\space \text{then} \\ \text{5 = B(1) (– 1)} \\ \Rightarrow \space \space B = -5 \\ \text{If x = 3, then} \\ 8 = C(2) (1) \\ \Rightarrow \space \space C = 4 \\ \therefore \space \int \frac{3x-1}{(x-1)(x-2)(x-3)}dx = \int \frac{dx}{x-1} + \int \frac{-5}{x-2}dx + \int \frac{4}{x-3}dx \\ \Rightarrow \space \space \int \frac{3x-1}{(x-1)(x-2)(x-3)}dx \\ = log\space |x-1|-5\space log\space |x-2| + 4\space log |x-3| +C. $$
$$\textbf{Ans. } \text{Let} \space \space I = \int ^{\pi/2}_0 \sqrt{sin\space \phi}\space cos^5 \phi\space d\phi \\ = \int ^{\pi/2}_0 \sqrt{sin \space \phi}\space cos^4\phi\space cos \phi \space d\phi \\ = \int ^{\pi/2}_0 \sqrt{sin\space \phi}(cos^2\space \phi)^2 cos\phi\space d\phi \\ \Rightarrow \space \space I = \space \int ^{\pi/2}_0 \sqrt{sin\space \phi}(1-sin^2\space \phi)^2\space cos\space\phi\space d\phi \\ \text{Put} \space \space sin\space \phi = t \\ \Rightarrow \space \space cos \space \phi \space d\phi = dt \\ \text{When} \space \phi = 0, \space t = sin\space 0 = 0\\ \text{when} \space \phi = \frac{\pi}{2},\space t = sin\space \frac{\pi}{2} \Rightarrow t = 1 \\ \therefore \space \space I = \int ^1_0 \sqrt{t}(1-t^2)^2 dt \\ = \int ^1_0 \sqrt{t}(1+t^4 – 2t^2)dt \\ =\int ^1_0(t^{1/2}+t^{9/2}-2t^{5/2})dt \\ =\begin{bmatrix}\frac{t^{3/2}}{3/2}+\frac{t^{11/2}}{11/2}-2\frac{t^{7/2}}{7/2} \end{bmatrix}^1_0 \\ = \begin{bmatrix}\frac{2}{3}\space t^{3/2}+\frac{2}{11}t^{11/2}-2×\frac{2}{7}t^{7/2} \end{bmatrix}^1_0 \\ = \space \left(\frac{2}{3} + \frac{2}{11}-\frac{4}{7} \right)-0 \\ = \frac{154+42-132}{231} \\ =\frac{196-132}{231} \\ = \frac{64}{231}$$
$$\textbf{Ans. } \text{Put }\space \space x = tan \space \theta \\ \Rightarrow \space \space \space dx \space = sec^2\space \theta \space d\theta \\ \text{and} \space \space \theta = tan^{-1}x \\ \text{When} \space x = \space 1, \space \theta = \tan^{-1} 1 = \frac{\pi}{4} \\ \text{When }\space x = \space 0,\space \theta = tan^{-1}0 = 0 \\ \therefore \space \space I =\int ^{\pi/4}_0\space sin^{-1}\left(\frac{2\space tan\space \theta}{1+tan^2\space \theta}\right).sec^2\space \theta\space d\theta \begin{bmatrix}\because sin\space 2A = \frac{2\space tan \space A}{1+tan^2 \space A} \end{bmatrix}\\ = \int^{\pi/4}_0 sin^{-1}(sin\space 2\theta)sec^2\space\theta\space d\theta \\ = \int^{\pi/4}_0 2\theta sec^2 \space\theta d\theta = 2\int ^{\pi/4}_0 {{\theta}\atop {I}}{{sec^2}\atop {II}}\theta \space d \theta \\ = 2 \begin{bmatrix}\theta\int^{\pi/4}_0 sec^2\theta \space d\theta-\int^{\pi/4}_0 1.tan\space \theta \space d\theta \end{bmatrix} \\ =2\left[\theta\space tan\theta + log |cos\theta|\right]^{\pi/4}_0 \\ = 2 \begin{Bmatrix}\frac{\pi}{4}tan\frac{\pi}{4} + log \space \left|cos\frac{\pi}{4} \right| \end{Bmatrix} -\{0+log\space 1 \} \\ = \space 2 \left[\frac{\pi}{4}+ log\frac{1}{\sqrt{2}}\right] \\ = \frac{\pi}{2} + 2\space log \space 2^{-1/2} \\= \frac{\pi}{2}-2 ×\frac{1}{2}\space log \space 2 \\ = \frac{\pi}{2}-\space log\space 2.$$
$$\textbf{Ans. } I =\space \int ^{\pi/2}_0 \frac{sin^{3/2}x}{sin^{3/2}x + cos^{3/2}x}dx \space \space \space \space …(i) \\ \text{We know that} \\ \int^a_0f(x) \space dx = \int^a_0 f(a-x)dx \\ \therefore \space \space I = \int^{\pi/2}_0 \frac{sin^{3/2}\left(\frac{\pi}{2}-x\right)}{sin^{3/2}\left(\frac{\pi}{2}-x\right)+cos^{3/2}\left(\frac{\pi}{2}-x\right)}dx \\ I = \int ^{\pi/2}_0 \frac{cos^{3/2}x}{cos^{3/2}x + sin^{3/2}x}dx \space \space \space \space …(ii) \\ \text{Adding equations (i) and (ii),} \\ 2I = \int ^{\pi/2}_0 \frac{sin^{3/2}x + cos^{3/2}x}{sin^{3/2}x + cos^{3/2}x}dx \\ \Rightarrow \space \space 2I = \int^{\pi/2}_0 1dx \\ =[x]^{\pi/2}_0 \\ \Rightarrow \space \space \space 2I = \frac{\pi}{2} \\ \Rightarrow \space \space I = \frac{\pi}{4}.$$
$$\textbf{Ans. } \text{We know that} \\ x +2 = 0 \\ x=-2 \\\int^5_{-5} f(x)\space dx = \int^c_af(x)dx + \int^b_c f(x)dx \\ \therefore \space \space I = \int^{-2}_{-5} -(x+2)dx + \int^5_{-2}(x+2)dx \\ =- \left[\frac{x^2}{2}+2x\right]^{-2}_{-5} + \left[\frac{x^2}{2} + 2x\right]^5_{-2} \\ = – \Bigg \lbrace \left(\frac{4}{2}-4\right)-\left(\frac{25}{2}-10\right) \Bigg \rbrace + \Bigg \lbrace \left(\frac{25}{2}+10\right)-\left(\frac{4}{2}-4\right) \Bigg \rbrace \\ =\space \frac{4}{2}+ \frac{5}{2}+\frac{45}{2}+\frac{4}{2}= \frac{58}{2} \\ \therefore \space \space \space I = 29.$$
$$\textbf{Ans. } \text{We know that } \\ \int^a_{-a} f(x)dx = \begin{cases} 0, if\space f(x)\space is \space odd \\ 2\int^a_0 f(x)dx, if\space f(x)\space is\space even \end{cases} \\ \text{Here} \space \space f(x) = sin^7x \\ f(-x) = sin^7(-x) \\ = -sin^7x \\ \therefore\space \space \space \text{f(x) is odd function.}\\ \therefore \space \space \int^{\pi/2}_{-\pi/2}sin^7 x \space dx = 0. $$
$$\textbf{Ans. }\text{Since, }\\ \int^{2a}_0 f(x)dx = \begin{cases} 0,\space \text{if f(x) is odd} \\ 2\int^a_0 f(x)dx, \space if\space f(x) \space is \space even \end{cases} \\ \text{We have,} \space \int^{2\pi}_0 cos^5 x\space dx \\ \text{Here} \space \space f(x) = cos^5x \\ \text{Then,} \space \space f(2\pi -x) = cos^5(2\pi-x) \\ = cos^5x \\ \therefore \space \space \text{f(x) is an even function. }\\ \therefore \space \space \int^{2\pi}_0 cos^5x\space dx = 2\int^\pi _0 cos^5 \space x dx \\ \text{Now,}\space \space f(\pi-x) = cos^5(\pi -x) \\ =-cos^5x \\ =-f(x) \\ \because \space cos^5x \space\text{ is odd} \\ \therefore \space \space \int ^{2\pi}_0 cos^5x\space dx = 2 × 0 = 0.$$
$$\textbf{Ans. }\int\frac{1}{x-x^3}\space dx= \space \int \frac{dx}{x(1-x^2)} \\ = \int \frac{dx}{x(1-x)(1+x)} \\ \frac{1}{x(1-x)(1+x)}=\frac{A}{x} + \frac{B}{1-x}+ \frac{C}{1+x} \\ 1 = A(1-x^2)+B(x)(1+x) + C(x) (1 – x) \\ \text{Put}\space x = 1, \text{then} \\ 1 = 2B \\ \Rightarrow \space B = \frac{1}{2} \\ \text{Put}\space x = 0, \space \text{then} \\ 1 = A(1) \\ \Rightarrow \space \space \space A =1 \\ \text{Put}\space x = – 1,\space \text{then} \\ 1 = C(-1)(1+1) \\ \Rightarrow \space \space -2\space C = 1 \\ \Rightarrow \space \space C = -\frac{1}{2} \\ I = \int \frac{dx}{x} + \frac{1}{2}\int\frac{dx}{1-x}-\frac{1}{2} \int \frac{dx}{1+x}\\ = log\space x + \frac{1}{2} \frac{log|1-x|}{-1}-\frac{1}{2}\space log\space (1+x) +C \\ = log\space x\space – \frac{1}{2} \{log (1 – x) + log (1 + x)\} + C \\ =\space log\space x – \frac{1}{2}\space log\space \{(1-x)(1+x)\} + C \\ I = log\space x – \frac{1}{2}\space log(1-x^2) + C $$
Ans. Rationalising the denominator $$ \\ I = \int^1_0 \frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}dx \\ = \int^1_0 \frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x})^2-(\sqrt{x})^2}dx \\ = \int^1_0\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} dx \\ =\int^1_0 (\sqrt{1+x}+\sqrt{x}) \space dx \\ = \left[\frac{(1+x)^{1/2+1}}{1/2+1}\right]^1_0 + \left[\frac{x^{1/2+1}}{1/2+1}\right]^1_0 \\ \\ = \frac{2}{3}\left[(1+x)^{3/2}+(x)^{3/2}\right]^1_0 \\ = \frac{2}{3}\left[(1+1)^{3/2}+(1)^{3/2}\right]-\frac{2}{3}[1+0] \\ = \frac{2}{3}(2^{3/2}-1) + \frac{2}{3} \\ = \frac{2}{3}.2^{3/2}-\frac{2}{3} + \frac{2}{3} \\ = \frac{2}{3}(8)^{1/2}=\frac{2}{3}\space.\space 2\sqrt{2} \\ I = \frac{4\sqrt{2}}{3}. $$