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

Use the power-reduction identity \sin^2\theta = \dfrac{1-\cos2\theta}{2} with \theta=2x+5: \sin^2(2x+5) = \dfrac{1-\cos(4x+10)}{2}.

Integrate term by term: \displaystyle\int \dfrac12\,dx = \dfrac{x}{2}, and \displaystyle\int \dfrac12\cos(4x+10)\,dx = \dfrac18\sin(4x+10).

Use the product-to-sum identity \sin A\cos B = \dfrac12[\sin(A+B)+\sin(A-B)] with A=3x,\,B=4x: \sin3x\cos4x = \dfrac12[\sin7x-\sin x].

Integrate term by term: \displaystyle\dfrac12\left[-\dfrac{\cos7x}{7}+\cos x\right].

First combine \cos4x\cos6x = \dfrac12[\cos10x+\cos2x], so the integrand is \dfrac12\cos2x\cos10x + \dfrac12\cos^2 2x.

Expand each piece: \cos2x\cos10x = \dfrac12[\cos12x+\cos8x] and \cos^2 2x = \dfrac{1+\cos4x}{2}.

Combining, the integrand equals \dfrac14\cos12x + \dfrac14\cos8x + \dfrac14\cos4x + \dfrac14.

Integrate term by term: each cosine term integrates to \dfrac{1}{4}\cdot\dfrac{\sin(k x)}{k} for its respective k, and the constant term gives \dfrac{x}{4}.

Use the triple-angle identity \sin^3\theta = \dfrac{3\sin\theta-\sin3\theta}{4} with \theta=2x+1: \sin^3(2x+1) = \dfrac{3\sin(2x+1)-\sin(6x+3)}{4}.

Integrate term by term: \displaystyle\dfrac34\int \sin(2x+1)\,dx = -\dfrac38\cos(2x+1), and \displaystyle\dfrac14\int \sin(6x+3)\,dx = -\dfrac{1}{24}\cos(6x+3) (subtracted).

Group as (\sin x\cos x)^3 and use \sin x\cos x = \dfrac{\sin2x}{2}, giving \sin^3x\cos^3x = \dfrac{\sin^3 2x}{8}.

Use the triple-angle identity \sin^3 2x = \dfrac{3\sin2x-\sin6x}{4}, so the integrand is \dfrac{1}{32}[3\sin2x-\sin6x].

Integrate term by term: \displaystyle\dfrac{3}{32}\int \sin2x\,dx = -\dfrac{3}{64}\cos2x, and \displaystyle\dfrac{1}{32}\int \sin6x\,dx = -\dfrac{1}{192}\cos6x (subtracted).

First combine \sin x\sin3x = \dfrac12[\cos2x-\cos4x].

Multiply by \sin2x: \dfrac12[\sin2x\cos2x - \sin2x\cos4x].

Using \sin2x\cos2x=\dfrac12\sin4x and \sin2x\cos4x=\dfrac12[\sin6x-\sin2x], this becomes \dfrac14\sin4x - \dfrac14\sin6x + \dfrac14\sin2x.

Integrate term by term, each contributing \dfrac14\cdot\left(-\dfrac{\cos kx}{k}\right) for its respective k.

Use the product-to-sum identity \sin A\sin B = \dfrac12[\cos(A-B)-\cos(A+B)] with A=4x,\,B=8x: \sin4x\sin8x = \dfrac12[\cos4x-\cos12x].

Integrate term by term: \displaystyle\dfrac12\left[\dfrac{\sin4x}{4}-\dfrac{\sin12x}{12}\right].

Use the half-angle identities 1-\cos x = 2\sin^2\left(\dfrac{x}{2}\right) and 1+\cos x = 2\cos^2\left(\dfrac{x}{2}\right).

So the integrand simplifies to \tan^2\left(\dfrac{x}{2}\right) = \sec^2\left(\dfrac{x}{2}\right)-1.

Integrate: \displaystyle\int \sec^2\left(\dfrac{x}{2}\right)dx = 2\tan\left(\dfrac{x}{2}\right), then subtract x for the -1 term.

Rewrite \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).

Integrate term by term: \displaystyle\int 1\,dx = x, and \displaystyle\dfrac12\int \sec^2\left(\dfrac{x}{2}\right)dx = \tan\left(\dfrac{x}{2}\right) (subtracted).

Write \sin^4x = \left(\dfrac{1-\cos2x}{2}\right)^2 = \dfrac14\left(1-2\cos2x+\cos^2 2x\right).

Use \cos^2 2x=\dfrac{1+\cos4x}{2} to simplify: \sin^4x = \dfrac38 - \dfrac12\cos2x + \dfrac18\cos4x.

Integrate term by term: \dfrac{3x}{8}, -\dfrac{\sin2x}{4} and \dfrac{\sin4x}{32}.

Write \cos^4 2x = \left(\dfrac{1+\cos4x}{2}\right)^2 = \dfrac14\left(1+2\cos4x+\cos^2 4x\right).

Use \cos^2 4x=\dfrac{1+\cos8x}{2} to simplify: \cos^4 2x = \dfrac38 + \dfrac12\cos4x + \dfrac18\cos8x.

Integrate term by term: \dfrac{3x}{8}, \dfrac{\sin4x}{8} and \dfrac{\sin8x}{64}.

Use \sin^2 x = 1-\cos^2 x = (1-\cos x)(1+\cos x), so \dfrac{\sin^2 x}{1+\cos x} = 1-\cos x.

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

Use the sum-to-product identity \cos A-\cos B = -2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) for both numerator (A=2x,B=2\alpha) and denominator (A=x,B=\alpha).

Numerator =-2\sin(x+\alpha)\sin(x-\alpha), denominator =-2\sin\left(\dfrac{x+\alpha}{2}\right)\sin\left(\dfrac{x-\alpha}{2}\right).

Expand \sin(x+\alpha) and \sin(x-\alpha) using the double-angle form, e.g. \sin(x+\alpha)=2\sin\left(\dfrac{x+\alpha}{2}\right)\cos\left(\dfrac{x+\alpha}{2}\right), and cancel the common sine factors.

This leaves 4\cos\left(\dfrac{x+\alpha}{2}\right)\cos\left(\dfrac{x-\alpha}{2}\right).

Apply the product-to-sum identity once more: 4\cos\left(\dfrac{x+\alpha}{2}\right)\cos\left(\dfrac{x-\alpha}{2}\right) = 2(\cos x+\cos\alpha), since the integrand simplifies to 2\cos x + 2\cos\alpha.

Note that 1+\sin2x = (\sin x+\cos x)^2, so the integrand is \dfrac{\cos x-\sin x}{(\sin x+\cos x)^2}.

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

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

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

Split \tan^3 2x\sec2x\,dx = \tan^2 2x\cdot(\tan2x\sec2x\,dx) = (\sec^2 2x-1)\cdot\dfrac{dt}{2}, and since t=\sec2x, this is (t^2-1)\dfrac{dt}{2}.

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

Write \tan^4 x = \tan^2 x(\sec^2 x-1) = \tan^2 x\sec^2 x - \sec^2 x + 1 (using \tan^2x=\sec^2x-1 once more on the last term).

Integrate each piece: put u=\tan x for \displaystyle\int \tan^2 x\sec^2 x\,dx = \dfrac{\tan^3 x}{3}; also \displaystyle\int \sec^2 x\,dx=\tan x and \displaystyle\int 1\,dx=x.

Split the fraction into two parts: \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}.

Rewrite as \sec x\tan x + \text{cosec}\,x\cot x.

Use the standard results \displaystyle\int \sec x\tan x\,dx = \sec x and \displaystyle\int \text{cosec}\,x\cot x\,dx = -\text{cosec}\,x.

Use \cos2x=1-2\sin^2 x, so \cos2x+2\sin^2 x = 1.

So the integrand simplifies to \dfrac{1}{\cos^2 x} = \sec^2 x.

Write \sin x\cos^3 x = \cos^4 x\tan x, so the integrand is \dfrac{\sec^4 x}{\tan x}.

Use \sec^4 x = (1+\tan^2 x)\sec^2 x, so the integrand becomes \dfrac{\sec^2 x}{\tan x} + \tan x\sec^2 x.

Put t=\tan x, dt=\sec^2 x\,dx: \displaystyle\int \dfrac{dt}{t} = \log|t|, and \displaystyle\int t\,dt = \dfrac{t^2}{2}.

Factor \cos2x = \cos^2x-\sin^2x = (\cos x-\sin x)(\cos x+\sin x), so the integrand becomes \dfrac{\cos x-\sin x}{\cos x+\sin x}.

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

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

Use the co-function identity \cos x = \sin\left(\dfrac{\pi}{2}-x\right), so \sin^{-1}(\cos x) = \dfrac{\pi}{2}-x over the relevant domain.

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

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

Expand the numerator using \sin[(x-a)-(x-b)] = \sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b), so the fraction splits into \dfrac{1}{\sin(b-a)}\left[\tan(x-a)-\tan(x-b)\right].

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

Split the fraction: \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.

Integrate term by term: \displaystyle\int \sec^2 x\,dx = \tan x, and \displaystyle\int \text{cosec}^2 x\,dx = -\cot x (subtracted, giving +\cot x).

Notice that \dfrac{d}{dx}(xe^x) = e^x + xe^x = e^x(1+x) — exactly the numerator.

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