Class 12 Maths NCERT Solutions Chapter 7 Ex 7.3 – Trigonometric Identities

Let I = \displaystyle\int \sin^2(2x+5)\,dx

\sin^2(2x+5) = \dfrac{1-\cos(4x+10)}{2} [∵ \sin^2\theta = \dfrac{1-\cos2\theta}{2}]

\displaystyle\int \sin^2(2x+5)\,dx = \int \dfrac{1-\cos(4x+10)}{2}\,dx = \dfrac12\int 1\,dx - \dfrac12\int \cos(4x+10)\,dx

= \dfrac{x}{2} - \dfrac12\cdot\dfrac{\sin(4x+10)}{4} [∵ \displaystyle\int \cos(ax+b)\,dx = \dfrac{1}{a}\sin(ax+b)]

Let I = \displaystyle\int \sin3x\cos4x\,dx

\sin3x\cos4x = \dfrac12[\sin(3x+4x)+\sin(3x-4x)] = \dfrac12[\sin7x-\sin x] [∵ \sin A\cos B = \dfrac12\{\sin(A+B)+\sin(A-B)\}]

\displaystyle\int \sin3x\cos4x\,dx = \dfrac12\int \sin7x\,dx - \dfrac12\int \sin x\,dx = \dfrac12\left[-\dfrac{\cos7x}{7}+\cos x\right]

Let I = \displaystyle\int \cos2x\cos4x\cos6x\,dx

We know \cos4x\cos6x = \dfrac12[\cos10x+\cos2x] ...(1) [∵ \cos A\cos B = \dfrac12\{\cos(A+B)+\cos(A-B)\}]

Using (1), \cos2x\cos4x\cos6x = \dfrac12\cos2x\cos10x + \dfrac12\cos^2 2x

Also, \cos2x\cos10x = \dfrac12[\cos12x+\cos8x] ...(2) [∵ \cos A\cos B = \dfrac12\{\cos(A+B)+\cos(A-B)\}], and \cos^2 2x = \dfrac{1+\cos4x}{2} ...(3) [∵ \cos^2\theta = \dfrac{1+\cos2\theta}{2}]

Using (2) and (3), \cos2x\cos4x\cos6x = \dfrac14\cos12x + \dfrac14\cos8x + \dfrac14\cos4x + \dfrac14

\therefore\ I = \dfrac14\displaystyle\int \cos12x\,dx + \dfrac14\int \cos8x\,dx + \dfrac14\int \cos4x\,dx + \dfrac14\int dx

Let I = \displaystyle\int \sin^3(2x+1)\,dx

\sin^3(2x+1) = \dfrac{3\sin(2x+1)-\sin(6x+3)}{4} [∵ \sin^3\theta = \dfrac14(3\sin\theta-\sin3\theta)]

\displaystyle\int \sin^3(2x+1)\,dx = \dfrac34\int \sin(2x+1)\,dx - \dfrac14\int \sin(6x+3)\,dx = -\dfrac38\cos(2x+1) + \dfrac{1}{24}\cos(6x+3)

Let I = \displaystyle\int \sin^3x\cos^3x\,dx

\sin^3x\cos^3x = (\sin x\cos x)^3 = \left(\dfrac{\sin2x}{2}\right)^3 = \dfrac{\sin^3 2x}{8} [∵ \sin x\cos x = \dfrac12\sin2x]

\sin^3 2x = \dfrac{3\sin2x-\sin6x}{4} [∵ \sin^3\theta = \dfrac14(3\sin\theta-\sin3\theta)], so \sin^3x\cos^3x = \dfrac{1}{32}[3\sin2x-\sin6x]

\displaystyle\int \sin^3x\cos^3x\,dx = \dfrac{3}{32}\int \sin2x\,dx - \dfrac{1}{32}\int \sin6x\,dx = -\dfrac{3}{64}\cos2x + \dfrac{1}{192}\cos6x

Let I = \displaystyle\int \sin x\sin2x\sin3x\,dx

We know \sin x\sin3x = \dfrac12[\cos2x-\cos4x] ...(1) [∵ \sin A\sin B = \dfrac12\{\cos(A-B)-\cos(A+B)\}]

Multiplying both sides of (1) by \sin2x, \sin x\sin2x\sin3x = \dfrac12[\sin2x\cos2x - \sin2x\cos4x]

Now \sin2x\cos2x=\dfrac12\sin4x ...(2) [∵ \sin A\cos B = \dfrac12\{\sin(A+B)+\sin(A-B)\}], and \sin2x\cos4x=\dfrac12[\sin6x-\sin2x] ...(3) [∵ \sin A\cos B = \dfrac12\{\sin(A+B)+\sin(A-B)\}]

Using (2) and (3), \sin x\sin2x\sin3x = \dfrac14\sin4x - \dfrac14\sin6x + \dfrac14\sin2x

\therefore\ I = \dfrac14\displaystyle\int \sin4x\,dx - \dfrac14\int \sin6x\,dx + \dfrac14\int \sin2x\,dx

Let I = \displaystyle\int \sin4x\sin8x\,dx

\sin4x\sin8x = \dfrac12[\cos(8x-4x)-\cos(8x+4x)] = \dfrac12[\cos4x-\cos12x] [∵ \sin A\sin B = \dfrac12\{\cos(A-B)-\cos(A+B)\}]

\displaystyle\int \sin4x\sin8x\,dx = \dfrac12\int \cos4x\,dx - \dfrac12\int \cos12x\,dx = \dfrac12\left[\dfrac{\sin4x}{4}-\dfrac{\sin12x}{12}\right]

Let I = \displaystyle\int \dfrac{1-\cos x}{1+\cos x}\,dx

1-\cos x = 2\sin^2\left(\dfrac{x}{2}\right) [∵ 1 - \cos\theta = 2\sin^2\left(\dfrac{\theta}{2}\right)], and 1+\cos x = 2\cos^2\left(\dfrac{x}{2}\right) [∵ 1 + \cos\theta = 2\cos^2\left(\dfrac{\theta}{2}\right)]

\dfrac{1-\cos x}{1+\cos x} = \tan^2\left(\dfrac{x}{2}\right) = \sec^2\left(\dfrac{x}{2}\right)-1 [∵ \tan^2\theta = \sec^2\theta-1]

I = \displaystyle\int \sec^2\left(\dfrac{x}{2}\right)dx - \int dx = 2\tan\left(\dfrac{x}{2}\right) - x

Let I = \displaystyle\int \dfrac{\cos x}{1+\cos x}\,dx

\dfrac{\cos x}{1+\cos x} = 1 - \dfrac{1}{1+\cos x} = 1 - \dfrac{1}{2\cos^2(x/2)} = 1 - \dfrac12\sec^2\left(\dfrac{x}{2}\right) [∵ 1 + \cos\theta = 2\cos^2\left(\dfrac{\theta}{2}\right)]

I = \displaystyle\int 1\,dx - \dfrac12\int \sec^2\left(\dfrac{x}{2}\right)dx = x - \tan\left(\dfrac{x}{2}\right)

Let I = \displaystyle\int \sin^4x\,dx

We know \sin^2x = \dfrac{1-\cos2x}{2} [∵ \sin^2\theta = \dfrac{1-\cos2\theta}{2}], so \sin^4x = \left(\dfrac{1-\cos2x}{2}\right)^2 = \dfrac14\left(1-2\cos2x+\cos^2 2x\right) ...(1)

Also, \cos^2 2x=\dfrac{1+\cos4x}{2} ...(2) [∵ \cos^2\theta = \dfrac{1+\cos2\theta}{2}]

Using (2) in (1), \sin^4x = \dfrac38 - \dfrac12\cos2x + \dfrac18\cos4x

\therefore\ I = \dfrac38\displaystyle\int dx - \dfrac12\int \cos2x\,dx + \dfrac18\int \cos4x\,dx

Let I = \displaystyle\int \cos^4 2x\,dx

\cos^4 2x = \left(\dfrac{1+\cos4x}{2}\right)^2 = \dfrac14\left(1+2\cos4x+\cos^2 4x\right) [∵ \cos^2\theta = \dfrac{1+\cos2\theta}{2}]

\cos^2 4x=\dfrac{1+\cos8x}{2} [same identity], so \cos^4 2x = \dfrac38 + \dfrac12\cos4x + \dfrac18\cos8x

\therefore\ I = \dfrac38\displaystyle\int dx + \dfrac12\int \cos4x\,dx + \dfrac18\int \cos8x\,dx

Let I = \displaystyle\int \dfrac{\sin^2 x}{1+\cos x}\,dx

\sin^2 x = 1-\cos^2 x = (1-\cos x)(1+\cos x) [∵ \sin^2\theta = 1-\cos^2\theta]

So \dfrac{\sin^2 x}{1+\cos x} = 1-\cos x

\displaystyle\int (1-\cos x)\,dx = x-\sin x

Let I = \displaystyle\int \dfrac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx

We know \cos2x-\cos2\alpha = -2\sin(x+\alpha)\sin(x-\alpha) ...(1) [∵ \cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)]

and \cos x-\cos\alpha = -2\sin\left(\dfrac{x+\alpha}{2}\right)\sin\left(\dfrac{x-\alpha}{2}\right) ...(2) [same identity]

From (1) and (2), \dfrac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha} = \dfrac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\left(\frac{x+\alpha}{2}\right)\sin\left(\frac{x-\alpha}{2}\right)} ...(3)

Also, \sin(x+\alpha)=2\sin\left(\dfrac{x+\alpha}{2}\right)\cos\left(\dfrac{x+\alpha}{2}\right) and \sin(x-\alpha)=2\sin\left(\dfrac{x-\alpha}{2}\right)\cos\left(\dfrac{x-\alpha}{2}\right) ...(4) [∵ \sin2\theta = 2\sin\theta\cos\theta]

Using (4) in (3), \dfrac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha} = 4\cos\left(\dfrac{x+\alpha}{2}\right)\cos\left(\dfrac{x-\alpha}{2}\right) = 2[\cos x+\cos\alpha] [∵ \cos A\cos B = \dfrac12\{\cos(A+B)+\cos(A-B)\}]

\therefore\ I = 2\displaystyle\int \cos x\,dx + 2\cos\alpha\int dx (treating \alpha as constant)

Let I = \displaystyle\int \dfrac{\cos x-\sin x}{1+\sin 2x}\,dx

1+\sin2x = \sin^2x+\cos^2x+2\sin x\cos x = (\sin x+\cos x)^2 [∵ \sin^2x+\cos^2x=1,\ \sin2x=2\sin x\cos x]

So the integrand is \dfrac{\cos x-\sin x}{(\sin x+\cos x)^2}

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

The integral becomes \displaystyle\int t^{-2}\,dt = -\dfrac{1}{t}

Let I = \displaystyle\int \tan^3 2x\sec 2x\,dx

Put \sec2x = t, so 2\sec2x\tan2x\,dx=dt

I = \displaystyle\int \tan^2 2x\,(\tan2x\sec2x\,dx) = \int (\sec^2 2x-1)\cdot\dfrac{dt}{2} [∵ \tan^2\theta = \sec^2\theta-1]

I = \dfrac12\displaystyle\int (t^2-1)\,dt = \dfrac12\left[\dfrac{t^3}{3}-t\right] = \dfrac{t^3}{6}-\dfrac{t}{2}

Let I = \displaystyle\int \tan^4 x\,dx

\tan^4 x = \tan^2 x\cdot\tan^2 x = \tan^2 x(\sec^2 x-1) [∵ \tan^2\theta = \sec^2\theta-1]

= \tan^2 x\sec^2 x - \tan^2 x = \tan^2 x\sec^2 x - (\sec^2 x-1) [∵ \tan^2\theta = \sec^2\theta - 1\text{, again}]

\tan^4 x = \tan^2 x\sec^2 x - \sec^2 x + 1

\therefore\ I = \displaystyle\int \tan^2 x\sec^2 x\,dx - \int \sec^2 x\,dx + \int dx

For the first integral, put \tan x = t, so \sec^2 x\,dx=dt:

\displaystyle\int \tan^2 x\sec^2 x\,dx = \int t^2\,dt = \dfrac{t^3}{3} = \dfrac{\tan^3 x}{3}

Also, \displaystyle\int \sec^2 x\,dx = \tan x and \displaystyle\int dx = x

Let I = \displaystyle\int \dfrac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\,dx

\dfrac{\sin^3 x}{\sin^2 x\cos^2 x} + \dfrac{\cos^3 x}{\sin^2 x\cos^2 x} = \dfrac{\sin x}{\cos^2 x} + \dfrac{\cos x}{\sin^2 x} = \sec x\tan x + \text{cosec}\,x\cot x [∵ \dfrac{\sin x}{\cos^2 x} = \sec x\tan x,\ \dfrac{\cos x}{\sin^2 x} = \text{cosec}\,x\cot x]

Using the standard results \displaystyle\int \sec x\tan x\,dx = \sec x and \displaystyle\int \text{cosec}\,x\cot x\,dx = -\text{cosec}\,x [∵ \dfrac{d}{dx}(\sec x) = \sec x\tan x,\ \dfrac{d}{dx}(\text{cosec}\,x) = -\text{cosec}\,x\cot x]

Let I = \displaystyle\int \dfrac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx

\cos2x=1-2\sin^2 x [∵ \cos2\theta = 1 - 2\sin^2\theta], so \cos2x+2\sin^2 x = 1

\dfrac{\cos2x+2\sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x

I = \displaystyle\int \sec^2 x\,dx = \tan x

Let I = \displaystyle\int \dfrac{1}{\sin x\cos^3 x}\,dx

Dividing the numerator and denominator by \cos^4 x, \dfrac{1}{\sin x\cos^3 x} = \dfrac{\sec^4 x}{\dfrac{\sin x}{\cos x}} = \dfrac{\sec^4 x}{\tan x}

\sec^4 x = \sec^2 x\cdot\sec^2 x = (1+\tan^2 x)\sec^2 x [∵ \sec^2\theta = 1+\tan^2\theta]

I = \displaystyle\int \dfrac{(1+\tan^2 x)\sec^2 x}{\tan x}\,dx

Put \tan x = t, so \sec^2 x\,dx=dt

I = \displaystyle\int \dfrac{1+t^2}{t}\,dt = \int \left(\dfrac{1}{t}+t\right)dt = \log|t| + \dfrac{t^2}{2}

Let I = \displaystyle\int \dfrac{\cos 2x}{(\cos x+\sin x)^2}\,dx

\cos2x = \cos^2x-\sin^2x = (\cos x-\sin x)(\cos x+\sin x) [∵ \cos2\theta = \cos^2\theta - \sin^2\theta], so the integrand becomes \dfrac{\cos x-\sin x}{\cos x+\sin x}

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

The integral becomes \displaystyle\int \dfrac{dt}{t} = \log|t|

Let I = \displaystyle\int \sin^{-1}(\cos x)\,dx

\cos x = \sin\left(\dfrac{\pi}{2}-x\right) [∵ \cos\theta = \sin\left(\dfrac{\pi}{2}-\theta\right)], so \sin^{-1}(\cos x) = \dfrac{\pi}{2}-x over the relevant domain

\displaystyle\int \left(\dfrac{\pi}{2}-x\right)dx = \dfrac{\pi}{2}x - \dfrac{x^2}{2}

Let I = \displaystyle\int \dfrac{1}{\cos(x-a)\cos(x-b)}\,dx

(x-a)-(x-b)=b-a, a constant. Multiplying and dividing by \sin(b-a):

I = \dfrac{1}{\sin(b-a)}\displaystyle\int \dfrac{\sin[(x-a)-(x-b)]}{\cos(x-a)\cos(x-b)}\,dx

\sin[(x-a)-(x-b)] = \sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b) [∵ \sin(A-B) = \sin A\cos B - \cos A\sin B]

I = \dfrac{1}{\sin(b-a)}\displaystyle\int \left[\tan(x-a)-\tan(x-b)\right]dx

= \dfrac{1}{\sin(b-a)}\Big[-\log|\cos(x-a)| + \log|\cos(x-b)|\Big] [∵ \displaystyle\int \tan\theta\,d\theta = -\log|\cos\theta|]

Let I = \displaystyle\int \dfrac{\sin^2 x-\cos^2 x}{\sin^2 x\cos^2 x}\,dx

\dfrac{\sin^2 x}{\sin^2 x\cos^2 x} - \dfrac{\cos^2 x}{\sin^2 x\cos^2 x} = \dfrac{1}{\cos^2 x} - \dfrac{1}{\sin^2 x} = \sec^2 x - \text{cosec}^2 x

I = \displaystyle\int \sec^2 x\,dx - \int \text{cosec}^2 x\,dx = \tan x - (-\cot x) = \tan x+\cot x [∵ \dfrac{d}{dx}(\cot x) = -\text{cosec}^2 x]

Let I = \displaystyle\int \dfrac{e^x(1+x)}{\cos^2(xe^x)}\,dx

\dfrac{d}{dx}(xe^x) = e^x + xe^x = e^x(1+x) [∵ product rule] — exactly the numerator

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