Class 12 Maths NCERT Solutions Chapter 7 Ex 7.10 – Properties of Definite Integrals

Use \cos^2x=\dfrac{1+\cos2x}{2}. The anti derivative \dfrac{x}{2}+\dfrac{\sin2x}{4} gives \dfrac{\pi}{4} after evaluating at the limits, since the sine terms vanish.

Let I denote the integral. By P_4 (replace x with \pi/2-x): I = \displaystyle\int_0^{\pi/2} \dfrac{\cos x}{\cos x+\sin x}\,dx.

Adding: 2I = \displaystyle\int_0^{\pi/2} 1\,dx = \dfrac{\pi}{2}.

Same trick as Q2: by P_4, the swapped form has the same denominator, so adding the two versions gives 2I = \displaystyle\int_0^{\pi/2} 1\,dx = \dfrac{\pi}{2}.

Again by P_4, the swapped form is \dfrac{\sin^5x}{\cos^5x+\sin^5x}, and adding gives 2I=\dfrac{\pi}{2}.

Split at x=-2: \displaystyle\int_{-5}^{-2} -(x+2)\,dx + \int_{-2}^5 (x+2)\,dx = 4.5+24.5.

Split at x=5: each piece is a triangular area of 4.5.

By P_4 (replace x with 1-x): \displaystyle\int_0^1 x(1-x)^n\,dx = \int_0^1 (1-x)x^n\,dx = \int_0^1 \left(x^n-x^{n+1}\right)dx = \dfrac{1}{n+1}-\dfrac{1}{n+2}.

Let I denote the integral. Since \tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}, P_4 gives 1+\tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{2}{1+\tan x}.

So I = \dfrac{\pi}{4}\log2 - I, giving 2I=\dfrac{\pi}{4}\log2.

By P_4: \displaystyle\int_0^2 x\sqrt{2-x}\,dx = \int_0^2 (2-x)\sqrt{x}\,dx = \int_0^2 \left(2\sqrt{x}-x^{3/2}\right)dx = \left[\dfrac43x^{3/2}-\dfrac25x^{5/2}\right]_0^2.

Split into 2\displaystyle\int_0^{\pi/2}\log\sin x\,dx - \int_0^{\pi/2}\log\sin2x\,dx. The first is the known standard result -\dfrac{\pi}{2}\log2.

For the second, put t=2x: \displaystyle\int_0^{\pi/2}\log\sin2x\,dx = \int_0^{\pi/2}\log\sin t\,dt = -\dfrac{\pi}{2}\log2 (using P_6, since \sin(\pi-t)=\sin t).

Combining both gives the answer.

Since \sin^2x is even, use P_7(i): \displaystyle\int_{-\pi/2}^{\pi/2} \sin^2x\,dx = 2\int_0^{\pi/2}\sin^2x\,dx = 2\times\dfrac{\pi}{4} (by the same symmetry as Q1).

Let I denote the integral. By P_4 (\sin(\pi-x)=\sin x): 2I = \pi\displaystyle\int_0^\pi \dfrac{dx}{1+\sin x} — call this integral J.

Using 1+\sin x=2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right), the anti derivative of \dfrac{1}{1+\sin x} is -\tan\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right), giving J=2.

So 2I=2\pi.

\sin^7x is odd, since \sin^7(-x)=-\sin^7x. By P_7(ii), the integral of an odd function over a symmetric interval is 0.

Since \cos^5(2\pi-x)=\cos^5x, use P_6 with a=\pi: \displaystyle\int_0^{2\pi}\cos^5x\,dx = 2\int_0^{\pi}\cos^5x\,dx.

Now apply P_4 to \displaystyle\int_0^{\pi}\cos^5x\,dx: since \cos(\pi-x)=-\cos x, this integral equals its own negative, so it must be 0.

Let I denote the integral. By P_4 (which swaps \sin x and \cos x, leaving the denominator unchanged): I = -I, so 2I=0.

Use 1+\cos x = 2\cos^2\left(\dfrac{x}{2}\right), so the integral becomes \pi\log2 + 2\displaystyle\int_0^{\pi}\log\cos\left(\dfrac{x}{2}\right)dx.

Putting t=\dfrac{x}{2} reduces the remaining integral to 2\displaystyle\int_0^{\pi/2}\log\cos t\,dt = -\pi\log2 (the same standard result as Q10). Combining: \pi\log2+2(-\pi\log2).

Let I denote the integral. By P_4: I = \displaystyle\int_0^a \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx. Adding: 2I = \displaystyle\int_0^a 1\,dx = a.

Split at x=1: \displaystyle\int_0^1 -(x-1)\,dx + \int_1^4 (x-1)\,dx = 0.5+4.5.

Let I=\displaystyle\int_0^a f(x)g(x)\,dx. By P_4 and the given f(a-x)=f(x): I = \displaystyle\int_0^a f(x)g(a-x)\,dx — call this (1).

Adding (1) to the original expression for I: 2I = \displaystyle\int_0^a f(x)\left[g(x)+g(a-x)\right]dx = 4\int_0^a f(x)\,dx, using the given condition g(x)+g(a-x)=4.

x^3, x\cos x, and \tan^5x are all odd functions, so their sum integrates to 0 over the symmetric interval by P_7(ii).

Only the constant term 1 survives, giving \displaystyle\int_{-\pi/2}^{\pi/2} 1\,dx = \pi.

Let I denote the integral. By P_4, swapping \sin x and \cos x gives I = \displaystyle\int_0^{\pi/2} \log\left(\dfrac{4+3\cos x}{4+3\sin x}\right)dx = -I (since \log(1/k)=-\log k).

So 2I=0.