Class 12 Maths NCERT Solutions Chapter 7 Miscellaneous Exercise – Integrals

Factor the denominator: x-x^3 = x(1-x)(1+x), then split into partial fractions \dfrac{1}{x(1-x)(1+x)} = \dfrac{A}{x}+\dfrac{B}{1-x}+\dfrac{C}{1+x}.

Putting x=0,1,-1 into 1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x) gives A=1, B=\dfrac12, C=-\dfrac12.

Integrate each term and combine the two log terms into a single logarithm of 1-x^2.

Rationalise by multiplying numerator and denominator by \sqrt{x+a}-\sqrt{x+b}; the denominator becomes (x+a)-(x+b)=a-b.

The integrand simplifies to \dfrac{\sqrt{x+a}-\sqrt{x+b}}{a-b}, which integrates term by term.

Put x=\dfrac{a}{t}, so dx=-\dfrac{a}{t^2}dt. Then ax-x^2 = \dfrac{a^2(t-1)}{t^2}, so \sqrt{ax-x^2}=\dfrac{a\sqrt{t-1}}{t} — call this (1).

Using (1), x\sqrt{ax-x^2} = \dfrac{a}{t}\cdot\dfrac{a\sqrt{t-1}}{t} = \dfrac{a^2\sqrt{t-1}}{t^2}, so the integrand becomes \dfrac{-a/t^2\,dt}{a^2\sqrt{t-1}/t^2} = -\dfrac{dt}{a\sqrt{t-1}}.

Integrate: -\dfrac1a\displaystyle\int (t-1)^{-1/2}\,dt = -\dfrac{2}{a}\sqrt{t-1}, then substitute t=\dfrac{a}{x} back, so t-1=\dfrac{a-x}{x}.

Factor x^4 out of the bracket: (x^4+1)^{3/4} = x^3(1+x^{-4})^{3/4}, so the integrand becomes x^{-5}(1+x^{-4})^{-3/4}.

Put t=1+x^{-4}, so dt=-4x^{-5}\,dx, giving x^{-5}dx=-\dfrac{dt}{4}.

The integral becomes -\dfrac14\displaystyle\int t^{-3/4}\,dt = -t^{1/4}.

Put x=t^6, so dx=6t^5\,dt, x^{1/2}=t^3, x^{1/3}=t^2. The integral becomes \displaystyle\int \dfrac{6t^5}{t^3+t^2}\,dt = 6\int \dfrac{t^3}{1+t}\,dt.

Divide: \dfrac{t^3}{1+t} = t^2-t+1-\dfrac{1}{1+t} — call this (1).

Integrating (1) term by term gives 6\left[\dfrac{t^3}{3}-\dfrac{t^2}{2}+t-\log|1+t|\right] = 2t^3-3t^2+6t-6\log|1+t|.

Substitute back t=x^{1/6}.

Write \dfrac{5x}{(x+1)(x^2+9)} = \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+9}, so 5x=A(x^2+9)+(Bx+C)(x+1).

Putting x=-1 gives A=-\dfrac12; comparing coefficients gives B=\dfrac12 and C=\dfrac92.

Integrate each piece: -\dfrac12\log|x+1|, then split \dfrac{x/2+9/2}{x^2+9} into \dfrac14\log(x^2+9) and \dfrac32\tan^{-1}\left(\dfrac{x}{3}\right).

Write \sin x = \sin[(x-a)+a] = \sin(x-a)\cos a+\cos(x-a)\sin a.

Dividing by \sin(x-a) gives \dfrac{\sin x}{\sin(x-a)} = \cos a + \sin a\cot(x-a).

Integrate term by term, using \displaystyle\int \cot(x-a)\,dx = \log|\sin(x-a)|.

Use e^{k\log x}=x^k to rewrite the integrand as \dfrac{x^5-x^4}{x^3-x^2}.

Factor: \dfrac{x^4(x-1)}{x^2(x-1)} = x^2 (for x\neq1).

Put t=\sin x, so dt=\cos x\,dx.

The integral becomes \displaystyle\int \dfrac{dt}{\sqrt{4-t^2}} = \sin^{-1}\left(\dfrac{t}{2}\right).

Factor the numerator: \sin^8x-\cos^8x = (\sin^4x-\cos^4x)(\sin^4x+\cos^4x) = -\cos2x\left(\sin^4x+\cos^4x\right), using \sin^2x-\cos^2x=-\cos2x.

Also \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x — exactly the denominator, so the two cancel.

The integrand simplifies entirely to -\cos2x.

Since (x+a)-(x+b)=a-b is constant, multiply and divide by \sin(a-b): \dfrac{1}{\cos(x+a)\cos(x+b)} = \dfrac{1}{\sin(a-b)}\cdot\dfrac{\sin[(x+a)-(x+b)]}{\cos(x+a)\cos(x+b)}.

Expanding the numerator using the sine-subtraction formula splits the fraction into \dfrac{1}{\sin(a-b)}\left[\tan(x+a)-\tan(x+b)\right].

Integrate using \displaystyle\int \tan\theta\,d\theta = -\log|\cos\theta| for each term.

Put t=x^4, so dt=4x^3\,dx.

The integral becomes \dfrac14\displaystyle\int \dfrac{dt}{\sqrt{1-t^2}} = \dfrac14\sin^{-1}t.

Put t=e^x, so dt=e^x\,dx. The integral becomes \displaystyle\int \dfrac{dt}{(1+t)(2+t)}.

Decompose: \dfrac{1}{(1+t)(2+t)} = \dfrac{1}{1+t}-\dfrac{1}{2+t}.

Write \dfrac{1}{(x^2+1)(x^2+4)} = \dfrac{A}{x^2+1}+\dfrac{B}{x^2+4}, so 1=A(x^2+4)+B(x^2+1).

Comparing coefficients gives A+B=0 and 4A+B=1, so A=\dfrac13, B=-\dfrac13.

Since e^{\log\sin x}=\sin x, the integrand is \cos^3x\sin x.

Put t=\cos x, so dt=-\sin x\,dx: the integral becomes -\displaystyle\int t^3\,dt = -\dfrac{t^4}{4}.

Since e^{3\log x}=x^3, the integrand is \dfrac{x^3}{x^4+1}.

Put t=x^4+1, so dt=4x^3\,dx: the integral becomes \dfrac14\displaystyle\int \dfrac{dt}{t} = \dfrac14\log|t|.

Put t=f(ax+b), so dt=a\,f'(ax+b)\,dx.

The integral becomes \dfrac1a\displaystyle\int t^n\,dt = \dfrac{1}{a}\cdot\dfrac{t^{n+1}}{n+1}.

Divide inside the square root by \sin^4x: \sin^3x\sin(x+\alpha) = \sin^4x\left[\cos\alpha+\sin\alpha\cot x\right] — call the bracket (1), using \dfrac{\sin(x+\alpha)}{\sin x}=\cos\alpha+\sin\alpha\cot x.

So the integrand becomes \dfrac{\text{cosec}^2x}{\sqrt{(1)}}, i.e. \dfrac{\text{cosec}^2x}{\sqrt{\cos\alpha+\sin\alpha\cot x}}.

Put u=\cos\alpha+\sin\alpha\cot x, so du=-\sin\alpha\,\text{cosec}^2x\,dx, giving \text{cosec}^2x\,dx=-\dfrac{du}{\sin\alpha}.

The integral becomes -\dfrac{1}{\sin\alpha}\displaystyle\int u^{-1/2}\,du = -\dfrac{2}{\sin\alpha}\sqrt{u}.

Multiply the fraction under the root by \dfrac{1-\sqrt x}{1-\sqrt x}: this gives \sqrt{\dfrac{(1-\sqrt x)^2}{1-x}} = \dfrac{1-\sqrt x}{\sqrt{1-x}} — call this (1).

Using (1), the integral splits as \displaystyle\int \dfrac{dx}{\sqrt{1-x}} - \int \dfrac{\sqrt x}{\sqrt{1-x}}\,dx. The first piece is -2\sqrt{1-x}.

For the second piece, substitute x=\sin^2\theta: \displaystyle\int \sqrt{\dfrac{x}{1-x}}\,dx = \int 2\sin^2\theta\,d\theta = \theta-\sin\theta\cos\theta = \sin^{-1}\sqrt{x}-\sqrt{x-x^2}.

Combining both pieces (the second is subtracted).

Use 1+\cos2x=2\cos^2x and \sin2x=2\sin x\cos x: \dfrac{2+\sin2x}{1+\cos2x} = \dfrac{1}{\cos^2x}+\dfrac{\sin x}{\cos x} = \sec^2x+\tan x.

This matches the form e^x[f(x)+f'(x)] with f(x)=\tan x, since f'(x)=\sec^2x.

Write \dfrac{x^2+x+1}{(x+1)^2(x+2)} = \dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x+2}, so x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2.

Putting x=-1 gives B=1; putting x=-2 gives C=3; comparing the x^2 coefficient gives A=-2.

Substitute x=\cos\theta: using the half-angle identity, \sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \tan\left(\dfrac{\theta}{2}\right).

So the integrand becomes \tan^{-1}\left[\tan\left(\dfrac{\theta}{2}\right)\right] = \dfrac{\theta}{2} = \dfrac12\cos^{-1}x.

Integrate \dfrac12\cos^{-1}x using the standard result \displaystyle\int \cos^{-1}x\,dx = x\cos^{-1}x-\sqrt{1-x^2}.

Rewrite \log(x^2+1)-2\log x = \log\left(\dfrac{x^2+1}{x^2}\right).

Put t=\dfrac{\sqrt{x^2+1}}{x}, so t^2=\dfrac{x^2+1}{x^2} and the log term equals 2\log t — call this (1).

Differentiating t: dt = -\dfrac{1}{x^3t}\,dx, so dx=-x^3t\,dt. Also \dfrac{\sqrt{x^2+1}}{x^4} = \dfrac{t}{x^3} — call this (2).

Combining (1) and (2) with dx, the integrand becomes \dfrac{t}{x^3}\cdot2\log t\cdot(-x^3t)\,dt = -2t^2\log t\,dt.

Integrate by parts with u=\log t,\,dv=t^2\,dt: \displaystyle\int t^2\log t\,dt = \dfrac{t^3}{3}\log t-\dfrac{t^3}{9}, so -2\displaystyle\int t^2\log t\,dt = -\dfrac{2t^3}{3}\log t+\dfrac{2t^3}{9}.

Substitute back t^3=\dfrac{(x^2+1)^{3/2}}{x^3} and \log t=\dfrac12\left[\log(x^2+1)-2\log x\right], then combine into a single fraction.

Use half-angle identities: 1-\cos x=2\sin^2\left(\dfrac{x}{2}\right) and \sin x=2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right), giving \dfrac{1-\sin x}{1-\cos x} = \dfrac12\text{cosec}^2\left(\dfrac{x}{2}\right)-\cot\left(\dfrac{x}{2}\right).

This matches f(x)+f'(x) with f(x)=-\cot\left(\dfrac{x}{2}\right), so the anti derivative is -e^x\cot\left(\dfrac{x}{2}\right).

At x=\pi, \cot\left(\dfrac{\pi}{2}\right)=0, giving 0.

At x=\dfrac{\pi}{2}, \cot\left(\dfrac{\pi}{4}\right)=1, giving -e^{\pi/2}.

Put t=\sin^2x, so dt=2\sin x\cos x\,dx, i.e. \sin x\cos x\,dx=\dfrac{dt}{2}. When x=0, t=0; when x=\dfrac{\pi}{4}, t=\dfrac12.

Also \cos^4x+\sin^4x = 1-2\sin^2x\cos^2x = 1-2t(1-t) = 2t^2-2t+1 — call this (1).

Using (1), the integral becomes \dfrac12\displaystyle\int_0^{1/2}\dfrac{dt}{2t^2-2t+1}. Complete the square: 2t^2-2t+1 = 2\left(t-\dfrac12\right)^2+\dfrac12.

This gives anti derivative \tan^{-1}(2t-1), evaluated from t=0 to t=\dfrac12: \dfrac12\left[\tan^{-1}(0)-\tan^{-1}(-1)\right] = \dfrac12\cdot\dfrac{\pi}{4}.

Divide numerator and denominator by \cos^2x: the integrand becomes \dfrac{1}{1+4\tan^2x}.

Put t=\tan x, so dx=\dfrac{dt}{1+t^2}; the limits become t=0 to t\to\infty.

The integral becomes \displaystyle\int_0^\infty \dfrac{dt}{(1+4t^2)(1+t^2)} — call this integral (1).

Decompose \dfrac{1}{(1+4t^2)(1+t^2)} = -\dfrac{1/3}{1+t^2}+\dfrac{4/3}{1+4t^2}.

Integrate (1) term by term using \displaystyle\int_0^\infty \dfrac{dt}{1+t^2}=\dfrac{\pi}{2} and \displaystyle\int_0^\infty \dfrac{dt}{1+4t^2}=\dfrac{\pi}{4}.

Combine: -\dfrac13\cdot\dfrac{\pi}{2}+\dfrac43\cdot\dfrac{\pi}{4} = -\dfrac{\pi}{6}+\dfrac{\pi}{3}.

Note (\sin x-\cos x)^2 = 1-\sin2x, so \sin2x = 1-(\sin x-\cos x)^2 — call this (1).

Put t=\sin x-\cos x, so dt=(\cos x+\sin x)\,dx. Using (1), the integral becomes \displaystyle\int \dfrac{dt}{\sqrt{1-t^2}} = \sin^{-1}t = \sin^{-1}(\sin x-\cos x).

Evaluate the limits: at x=\dfrac{\pi}{3}, t=\dfrac{\sqrt3-1}{2}; at x=\dfrac{\pi}{6}, t=-\dfrac{\sqrt3-1}{2}.

Since \sin^{-1} is odd, \sin^{-1}\left(\dfrac{\sqrt3-1}{2}\right)-\sin^{-1}\left(-\dfrac{\sqrt3-1}{2}\right) = 2\sin^{-1}\left(\dfrac{\sqrt3-1}{2}\right).

Rationalise by multiplying numerator and denominator by \sqrt{1+x}+\sqrt{x}; the denominator becomes (1+x)-x=1.

The integral becomes \displaystyle\int_0^1 \left(\sqrt{1+x}+\sqrt{x}\right)dx = \left[\dfrac23(1+x)^{3/2}+\dfrac23x^{3/2}\right]_0^1.

At x=1: \dfrac23(2\sqrt2)+\dfrac23 = \dfrac{4\sqrt2}{3}+\dfrac23. At x=0: \dfrac23.

Using the identity from Q27, \sin2x = 1-(\sin x-\cos x)^2, so 9+16\sin2x = 25-16(\sin x-\cos x)^2 — call this (1).

Put t=\sin x-\cos x, so dt=(\cos x+\sin x)\,dx. Using (1), the integral becomes \displaystyle\int \dfrac{dt}{25-16t^2} = \dfrac{1}{40}\log\left|\dfrac{5+4t}{5-4t}\right|.

Evaluate the limits: at x=\dfrac{\pi}{4}, t=0; at x=0, t=-1.

Combine: \dfrac{1}{40}\left[\log1-\log\left|\dfrac19\right|\right] = \dfrac{1}{40}\log9 = \dfrac{1}{20}\log3.

Take u=\tan^{-1}(\sin x) and dv=\sin2x\,dx=2\sin x\cos x\,dx; choosing v=\sin^2x avoids introducing \cos2x.

By parts: \displaystyle\int_0^{\pi/2} \sin2x\tan^{-1}(\sin x)\,dx = \left[\sin^2x\tan^{-1}(\sin x)\right]_0^{\pi/2} - \int_0^{\pi/2}\dfrac{\sin^2x\cos x}{1+\sin^2x}\,dx.

The boundary term evaluates to \dfrac{\pi}{4}-0=\dfrac{\pi}{4} — call this (1).

For the remaining integral, put t=\sin x: \displaystyle\int_0^1 \dfrac{t^2}{1+t^2}\,dt = \int_0^1\left(1-\dfrac{1}{1+t^2}\right)dt = 1-\dfrac{\pi}{4} — call this (2).

Combine (1) and (2): \dfrac{\pi}{4}-\left(1-\dfrac{\pi}{4}\right) = \dfrac{\pi}{2}-1.

Split into three separate integrals. Since x\ge1 throughout, \displaystyle\int_1^4|x-1|\,dx = \int_1^4(x-1)\,dx = 4.5.

For |x-2|, split at x=2: \displaystyle\int_1^2-(x-2)\,dx+\int_2^4(x-2)\,dx = 0.5+2 = 2.5.

For |x-3|, split at x=3: \displaystyle\int_1^3-(x-3)\,dx+\int_3^4(x-3)\,dx = 2+0.5 = 2.5.

Add all three results: 4.5+2.5+2.5.

Write \dfrac{1}{x^2(x+1)} = \dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x+1}.

Putting x=0 gives B=1; putting x=-1 gives C=1; comparing the x^2 coefficient gives A=-1.

Anti derivative: -\log|x|-\dfrac1x+\log|x+1|.

At x=3: -\log3-\dfrac13+\log4. At x=1: -1+\log2.

Subtracting: \left(-\log3-\dfrac13+\log4\right)-(-1+\log2) = \dfrac23+\log\left(\dfrac{4}{3\cdot2}\right) = \dfrac23+\log\left(\dfrac23\right), which is the required result.

By integration by parts (u=x,\,dv=e^x\,dx): \displaystyle\int xe^x\,dx = xe^x-e^x.

Evaluate: at x=1, e-e=0; at x=0, 0-1=-1.

Subtracting: 0-(-1)=1, which is the required result.

Let f(x)=x^{17}\cos^4x. Then f(-x) = (-x)^{17}\cos^4(-x) = -x^{17}\cos^4x = -f(x), so f is an odd function.

By property P_7(ii), the integral of an odd function over a symmetric interval [-a,a] is 0.

Use the triple-angle identity \sin^3x = \dfrac{3\sin x-\sin3x}{4}.

Integrate: \displaystyle\int_0^{\pi/2}3\sin x\,dx = \left[-3\cos x\right]_0^{\pi/2} = 3, and \displaystyle\int_0^{\pi/2}\sin3x\,dx = \left[-\dfrac{\cos3x}{3}\right]_0^{\pi/2} = \dfrac13.

Combine: \dfrac14\left(3-\dfrac13\right) = \dfrac14\cdot\dfrac83, which equals the required result.

Write 2\tan^3x = 2\tan x\left(\sec^2x-1\right) = 2\tan x\sec^2x-2\tan x.

Anti derivative: \displaystyle\int 2\tan x\sec^2x\,dx = \tan^2x (put u=\tan x), and \displaystyle\int 2\tan x\,dx = -2\log|\cos x|, giving overall \tan^2x+2\log|\cos x|.

Evaluate: at x=\dfrac{\pi}{4}, \tan^2\left(\dfrac{\pi}{4}\right)+2\log\left(\dfrac{1}{\sqrt2}\right) = 1-\log2; at x=0, this equals 0.

By integration by parts (u=\sin^{-1}x,\,dv=dx): \displaystyle\int \sin^{-1}x\,dx = x\sin^{-1}x+\sqrt{1-x^2} (since \displaystyle\int \dfrac{-x}{\sqrt{1-x^2}}\,dx = \sqrt{1-x^2}).

Evaluate: at x=1, 1\cdot\dfrac{\pi}{2}+0 = \dfrac{\pi}{2}; at x=0, 0+1=1.

Subtracting gives the required result.

Multiply numerator and denominator by e^x: the integrand becomes \dfrac{e^x}{e^{2x}+1}.

Put t=e^x, so dt=e^x\,dx: the integral becomes \displaystyle\int \dfrac{dt}{t^2+1} = \tan^{-1}t.

Factor \cos2x = (\cos x-\sin x)(\cos x+\sin x), so the integrand simplifies to \dfrac{\cos x-\sin x}{\cos x+\sin x}.

Put t=\sin x+\cos x, so dt=(\cos x-\sin x)\,dx — exactly the numerator.

Let I=\displaystyle\int_a^b x\,f(x)\,dx. By P_3, replacing x with a+b-x: I = \displaystyle\int_a^b (a+b-x)\,f(a+b-x)\,dx.

Since f(a+b-x)=f(x), this becomes I = \displaystyle\int_a^b (a+b-x)\,f(x)\,dx = (a+b)\int_a^b f(x)\,dx - I.

So 2I=(a+b)\displaystyle\int_a^b f(x)\,dx, giving I=\dfrac{a+b}{2}\displaystyle\int_a^b f(x)\,dx.