Class 12 Ch 7: Integrals - Exercise 7.10 Practice

Q1

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Evaluate the integral: $\int_0^{\pi/2} \cos^2 x dx$

Let $I = \int_0^{\pi/2} \cos^2 x dx$. Using $\int_0^a f(x)dx = \int_0^a f(a-x)dx$:

$I = \int_0^{\pi/2} \cos^2(\frac{\pi}{2}-x) dx = \int_0^{\pi/2} \sin^2 x dx$.

Adding both: $2I = \int_0^{\pi/2} (\cos^2 x + \sin^2 x) dx = \int_0^{\pi/2} 1 dx$.

$2I = [x]_0^{\pi/2} = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}$.

$\boxed{\frac{\pi}{4}}$

Q2

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Evaluate the integral: $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.

Using property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$:

$I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.

Adding: $2I = \int_0^{\pi/2} 1 dx = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}$.

$\boxed{\frac{\pi}{4}}$

Q3

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Evaluate the integral: $\int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx$

Similar to Q2, using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx$.

$2I = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$.

$\boxed{\frac{\pi}{4}}$

Q4

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Evaluate the integral: $\int_0^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} dx$

Using the same property as above.

$I = \int_0^{\pi/2} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} dx$.

$2I = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$.

$\boxed{\frac{\pi}{4}}$

Q5

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Evaluate the integral: $\int_{-5}^5 |x+2| dx$

$|x+2|$ changes sign at $x=-2$. Split integral at $-2$.

$\int_{-5}^{-2} -(x+2) dx + \int_{-2}^5 (x+2) dx$.

$[-\frac{x^2}{2}-2x]_{-5}^{-2} + [\frac{x^2}{2}+2x]_{-2}^5$.

$= 4.5 + 24.5 = 29$.

$\boxed{29}$

Q6

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Evaluate the integral: $\int_2^8 |x-5| dx$

Split at $x=5$. $\int_2^5 -(x-5) dx + \int_5^8 (x-5) dx$.

$[5x-\frac{x^2}{2}]_2^5 + [\frac{x^2}{2}-5x]_5^8$.

$= 4.5 + 4.5 = 9$.

$\boxed{9}$

Q7

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Evaluate the integral: $\int_0^1 x(1-x)^n dx$

Use $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^1 (1-x)(1-(1-x))^n dx = \int_0^1 (1-x)x^n dx$.

$= \int_0^1 (x^n - x^{n+1}) dx = [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}]_0^1$.

$\boxed{\frac{1}{(n+1)(n+2)}}$

Q8

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Evaluate the integral: $\int_0^{\pi/4} \log(1+\tan x) dx$

$I = \int_0^{\pi/4} \log(1+\tan(\frac{\pi}{4}-x)) dx = \int_0^{\pi/4} \log(1+\frac{1-\tan x}{1+\tan x}) dx$.

$= \int_0^{\pi/4} \log(\frac{2}{1+\tan x}) dx = \int_0^{\pi/4} (\log 2 - \log(1+\tan x)) dx$.

$I = \frac{\pi}{4}\log 2 - I \Rightarrow 2I = \frac{\pi}{4}\log 2$.

$\boxed{\frac{\pi}{8}\log 2}$

Q9

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Evaluate the integral: $\int_0^2 x\sqrt{2-x} dx$

Use property: $\int_0^2 (2-x)\sqrt{2-(2-x)} dx = \int_0^2 (2-x)\sqrt{x} dx$.

$= \int_0^2 (2x^{1/2} - x^{3/2}) dx = [\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}]_0^2$.

$= \frac{4}{3}(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) = 8\sqrt{2}(\frac{1}{3} - \frac{1}{5})$.

$\boxed{\frac{16\sqrt{2}}{15}}$

Q10

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Evaluate the integral: $\int_0^{\pi/2} (2\log \sin x - \log \sin 2x) dx$

Simplify integrand: $\log(\sin^2 x) - \log(2\sin x \cos x) = \log(\frac{\sin x}{2\cos x}) = \log(\frac{1}{2}\tan x)$.

$I = \int_0^{\pi/2} (\log \tan x - \log 2) dx$.

$\int_0^{\pi/2} \log \tan x dx = 0$ (Property).

So $I = -\frac{\pi}{2}\log 2 = \frac{\pi}{2}\log(\frac{1}{2})$.

$\boxed{\frac{\pi}{2}\log \frac{1}{2}}$

Q11

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Evaluate the integral: $\int_{-\pi/2}^{\pi/2} \sin^2 x dx$

Let $f(x) = \sin^2 x$. Since $f(-x) = (\sin(-x))^2 = (-\sin x)^2 = \sin^2 x = f(x)$, it is an even function.

Using property $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$:

$I = 2\int_0^{\pi/2} \sin^2 x dx = 2\int_0^{\pi/2} \frac{1-\cos 2x}{2} dx$.

$= \int_0^{\pi/2} (1-\cos 2x) dx = [x - \frac{\sin 2x}{2}]_0^{\pi/2}$.

$= (\frac{\pi}{2} - 0) - 0 = \frac{\pi}{2}$.

$\boxed{\frac{\pi}{2}}$

Q12

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Evaluate the integral: $\int_0^\pi \frac{x dx}{1+\sin x}$

Let $I = \int_0^\pi \frac{x}{1+\sin x} dx$. Use $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^\pi \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_0^\pi \frac{\pi-x}{1+\sin x} dx$.

Adding: $2I = \int_0^\pi \frac{\pi}{1+\sin x} dx = \pi \int_0^\pi \frac{1-\sin x}{\cos^2 x} dx$.

$2I = \pi \int_0^\pi (\sec^2 x - \sec x \tan x) dx = \pi [\tan x - \sec x]_0^\pi$.

$2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)] = \pi [(0 - (-1)) - (0 - 1)] = 2\pi$.

$\boxed{\pi}$

Q13

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Evaluate the integral: $\int_{-\pi/2}^{\pi/2} \sin^7 x dx$

Let $f(x) = \sin^7 x$.

$f(-x) = (\sin(-x))^7 = (-\sin x)^7 = -\sin^7 x = -f(x)$.

Since $f(x)$ is an odd function, $\int_{-a}^a f(x) dx = 0$.

$\boxed{0}$

Q14

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Evaluate the integral: $\int_0^{2\pi} \cos^5 x dx$

Let $f(x) = \cos^5 x$. $f(2\pi-x) = \cos^5 x$. So $I = 2\int_0^\pi \cos^5 x dx$.

Now check $f(\pi-x) = \cos^5(\pi-x) = (-\cos x)^5 = -\cos^5 x = -f(x)$.

Using property $\int_0^{2a} f(x) dx = 0$ if $f(2a-x) = -f(x)$ (here limit is $\pi$).

So $\int_0^\pi \cos^5 x dx = 0$.

$\boxed{0}$

Q15

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Evaluate the integral: $\int_0^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x} dx$

Use property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^{\pi/2} \frac{\cos x - \sin x}{1+\cos x \sin x} dx = - \int_0^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x} dx = -I$.

$2I = 0 \Rightarrow I = 0$.

$\boxed{0}$

Q16

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Evaluate the integral: $\int_0^\pi \log(1+\cos x) dx$

$I = \int_0^\pi \log(1+\cos(\pi-x)) dx = \int_0^\pi \log(1-\cos x) dx$.

$2I = \int_0^\pi \log(1-\cos^2 x) dx = \int_0^\pi \log(\sin^2 x) dx = 2\int_0^\pi \log \sin x dx$.

$I = \int_0^\pi \log \sin x dx = 2\int_0^{\pi/2} \log \sin x dx$.

Using standard result $\int_0^{\pi/2} \log \sin x dx = -\frac{\pi}{2}\log 2$.

$I = 2(-\frac{\pi}{2}\log 2) = -\pi \log 2$.

$\boxed{-\pi \log 2}$

Q17

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Evaluate the integral: $\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$

Use property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx$.

Adding: $2I = \int_0^a 1 dx = [x]_0^a = a$.

$\boxed{\frac{a}{2}}$

Q18

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Evaluate the integral: $\int_0^4 |x-1| dx$

Split at $x=1$: $\int_0^1 -(x-1) dx + \int_1^4 (x-1) dx$.

$[x - \frac{x^2}{2}]_0^1 + [\frac{x^2}{2} - x]_1^4$.

$= (1 - 0.5) + [(8-4) - (0.5-1)] = 0.5 + 4 + 0.5 = 5$.

$\boxed{5}$

Q19

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Show that $\int_0^a f(x) g(x) dx = 2 \int_0^a f(x) dx$, if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

$I = \int_0^a f(a-x)g(a-x) dx = \int_0^a f(x)(4-g(x)) dx$.

$I = 4\int_0^a f(x) dx - \int_0^a f(x)g(x) dx$.

$I = 4\int_0^a f(x) dx - I \Rightarrow 2I = 4\int_0^a f(x) dx$.

$\boxed{I = 2\int_0^a f(x) dx}$

Q20

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The value of $\int_{-\pi/2}^{\pi/2} (x^3 + x \cos x + \tan^5 x + 1) dx$ is:
(A) 0
(B) 2
(C) $\pi$
(D) 1

Let $f(x) = x^3 + x \cos x + \tan^5 x + 1$.

$x^3, x \cos x, \tan^5 x$ are odd functions. Their integral over symmetric limits is 0.

Remaining part: $\int_{-\pi/2}^{\pi/2} 1 dx = [x]_{-\pi/2}^{\pi/2} = \pi$.

$\boxed{\text{(C) } \pi}$

Q21

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The value of $\int_0^{\pi/2} \log\left(\frac{4+3\sin x}{4+3\cos x}\right) dx$ is:
(A) 2
(B) 3/4
(C) 0
(D) -2

Use property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$.

$I = \int_0^{\pi/2} \log\left(\frac{4+3\cos x}{4+3\sin x}\right) dx = \int_0^{\pi/2} -\log\left(\frac{4+3\sin x}{4+3\cos x}\right) dx = -I$.

$2I = 0 \Rightarrow I = 0$.

$\boxed{\text{(C) } 0}$