Integrals Test

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Integrals Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Evaluate $$\displaystyle\int^{\pi/4}_0\tan^2xdx$$

A
$$\left(1-\dfrac{\pi}{4}\right)$$

B
$$\left(1+\dfrac{\pi}{4}\right)$$

C
$$\left(1-\dfrac{\pi}{2}\right)$$

D
$$\left(1+\dfrac{\pi}{2}\right)$$

Solution
Question 2
1 / -0

Evaluate$$\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$$

A
$$0$$

B
$$\dfrac{\pi}{4}$$

C
$$e^{\pi/2}$$

D
$$(e^{\pi/2}-1)$$

Solution

Let $$I=\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$$

$$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{(1+\cos x)}+\dfrac{\sin x}{(1+\cos x)}\right\}dx$$

$$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{2\cos^2(x/2)}+\dfrac{2\sin(x/2)\cos (x/2)}{2\cos^2(x/2)}\right\}dx$$

$$=\displaystyle\int^{\pi/2}_0e^x\left\{\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right\}dx$$

$$=\left[e^x\tan \dfrac{x}{2}\right]^{\pi/2}_0=e^{\pi/2}$$.
Question 3
1 / -0

Evaluate $$\displaystyle\int^{\pi}_0\dfrac{dx}{(1+\sin x)}$$

A
$$\dfrac{1}{2}$$

B
$$1$$

C
$$2$$

D
$$0$$

Solution

Let $$I=\displaystyle\int^{\pi}_0\dfrac{dx}{(1+\sin x)}$$

$$=\displaystyle\int^{\pi}_0\dfrac{(1-\sin x)}{(1-\sin^2x)}dx$$

$$=\displaystyle\int^{\pi}_0\dfrac{(1-\sin x)}{\cos^2x}dx$$.

$$=\displaystyle\int^{\pi}_0[\sec^2x-\sec x\tan x]dx$$

$$=[\tan x-\sec x]^{\pi}_0$$

$$=(0+1)-(0-1)$$
$$=2$$.
Question 4
1 / -0

Evaluate : $$\displaystyle\int^9_0\dfrac{dx}{(1+\sqrt{x})}$$

A
$$(3-2 log 2)$$

B
$$(3+2 log 2)$$

C
$$(6-2 log 4)$$

D
$$(6+2 log 4)$$

Solution
Question 5
1 / -0

Evaluate $$\displaystyle\int^1_0\dfrac{xe^x}{(1+x)^2}dx$$

A
$$\left(\dfrac{e}{2}-1\right)$$

B
$$(e-1)$$

C
$$e(e-1)$$

D
None of these

Solution

Let $$I=\displaystyle\int^1_0\dfrac{xe^x}{(1+x)^2}dx$$

$$=\displaystyle\int^1_0e^x\left\{\dfrac{(1+x)-1}{(1+x)^2}\right\}dx$$

$$=\displaystyle\int^1_0e^x\left\{\dfrac{1}{(1+x)}-\dfrac{1}{(1+x)^2}\right\}dx$$

$$=\displaystyle\int^1_0e^x\{f(x)+f'(x)dx,$$ where $$f(x)=\dfrac{1}{(1+x)}$$

$$=[e^xf(x)]^1_0$$

$$=\left[e^x\cdot \dfrac{1}{(1+x)}\right]^1_0$$

$$=\left(\dfrac{e}{2}-1\right)$$.
Question 6
1 / -0

Evaluate $$\displaystyle\int^{\pi/2}_{\pi/4}\cot xdx$$

A
$$log 2$$

B
$$2 log 2$$

C
$$\dfrac{1}{2}log 2$$

D
None of these

Solution
Question 7
1 / -0

Evaluate $$\displaystyle\int^{\pi/2}_{\pi/3} cosec xdx$$

A
$$\dfrac{1}{2} log 2$$

B
$$\dfrac{1}{2} log 3$$

C
$$-log 2$$

D
None of these

Solution
Question 8
1 / -0

Evaluate $$\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$$

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{4}$$

C
$$\pi$$

D
None of these

Solution

Let $$I=\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$$

Put $$\sin x=t$$ and $$\cos xdx=dt$$.

$$[x=0\Rightarrow t=0]$$ and $$\left[x=\dfrac{\pi}{2}\Rightarrow t=1\right]$$.

$$\therefore I=\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$$
$$=[\tan^{-1}t]^1_0$$ ....... $$\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C$$

$$=\dfrac{\pi}{4}-0\\$$.
$$=\dfrac{\pi}{4}$$.
Question 9
1 / -0

Evaluate $$\displaystyle\int^{1}_0\dfrac{(1-x)}{(1+x)}dx$$

A
$$\dfrac{1}{2}log 2$$

B
$$(2log 2+1)$$

C
$$(2log 2-1)$$

D
$$\left(\dfrac{1}{2}log 2-1\right)$$

Solution
Question 10
1 / -0

Evaluate $$\displaystyle\int^{\pi/2}_0\cos^3xdx $$

A
$$1$$

B
$$\dfrac{3}{4}$$

C
$$\dfrac{2}{3}$$

D
None of these

Solution