MCQ 512 Marks
$\int_{-1 / 2}^{1 / 2}(\cos x)\left[\log \left(\frac{1-x}{1+x}\right)\right] d x=$
- ✓
$0$
- B
- C
$e^{\frac{1}{2}}$
- D
$2 e ^{\frac{1}{2}}$
Answer(A)
Let $f(x)=\cos x \log \left(\frac{1-x}{1+x}\right)$
$\therefore \quad f(-x)=\cos x \log \left(\frac{1-x}{1+x}\right)^{-1}$
$=-\cos x \log \left(\frac{1-x}{1+x}\right)=-f(x)$
$\begin{array}{ll}\therefore & f(x) \text { is an odd function. } \\ \therefore & \int_{-\frac{1}{2}}^{\frac{1}{2}} \cos x \log \left(\frac{1-x}{1+x}\right) d x=0\end{array}$
View full question & answer→MCQ 522 Marks
$\int_{-1}^1\left( e ^{x^3}+ e ^{-x^3}\right)\left( e ^x- e ^{-x}\right) d x$ is equal to
- A
$\frac{ e ^2}{2}-2 e$
- B
$e^2-2 e$
- ✓
$0$
- D
$2 e ^{-2}-2 e$
Answer(C)
Let $f (x)=\left( e ^{x^3}+ e ^{-x^3}\right)\left( e ^x- e ^{-x}\right)$
$\therefore \quad f (-x)=\left( e ^{-x^3}+ e ^{x^3}\right)\left( e ^{-x}- e ^x\right)$
$=-\left( e ^{x^3}+ e ^{-x^3}\right)\left( e ^x- e ^{-x}\right)=- f (x)$
$\therefore f (x)$ is an odd function.
$\therefore \quad \int_{-1}^1\left(e^{x^3}+e^{-x^3}\right)\left(e^x-e^{-x}\right) d x=0$
View full question & answer→MCQ 532 Marks
The value of $\int_{-1}^1\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) d x$ is
Answer(A)
Let $f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$
$\begin{array}{ll}\therefore & f (-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2}=- f (x) \\ \therefore & f (x) \text { is an odd function. } \\ \therefore & \int_{-1}^1 f (x) d x=0\end{array}$
View full question & answer→MCQ 542 Marks
$\int_{-a}^a \sin x f(\cos x) d x=$
Answer(B)
Let $f (x)=\sin x f (\cos x)$
$\begin{array}{ll}\therefore & f (-x)=-\sin x f (\cos x)=- f (x) \\ \therefore & f (x) \text { is an odd function. } \\ \therefore & \int_{- a }^{ a } f (x) d x=0\end{array}$
View full question & answer→MCQ 552 Marks
$\int_0^\pi x \log \sin x d x=$
AnswerCorrect option: B. $\frac{\pi^2}{2} \log \frac{1}{2}$
(B)
Let $I =\int_0^\pi x \log \sin x d x$ ...(i)
$\therefore \quad I=\int_0^\pi(\pi-x) \log \sin x d x$ ...(ii)
$\ldots \left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \log \sin x d x=2 \pi \int_0^{\frac{\pi}{2}} \log \sin x d x$
$\int_0^{\pi / 2} \log (\sin x) d x=\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2} \log 2$
$\therefore \quad 2 I =2 \pi\left(-\frac{\pi}{2} \log 2\right)$
$\Rightarrow I=\pi\left(\frac{\pi}{2} \log \frac{1}{2}\right)=\frac{\pi^2}{2} \log \frac{1}{2}$
View full question & answer→MCQ 562 Marks
$\int_0^{\frac{\pi}{2}} \log \sin x d x=$
AnswerCorrect option: A. $-\left(\frac{\pi}{2}\right) \log 2$
(A)
Let $I =\int_0^{\frac{\pi}{2}} \log \sin x d x$
$=\int_0^{\frac{\pi}{4}}(\log \sin x+\log \cos x) d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$\begin{aligned}= & \int_0^{\frac{\pi}{4}} \log \sin x \cos x d x \\ = & \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin 2 x}{2}\right) d x \\ = & \int_0^{\frac{\pi}{4}} \log \sin 2 x d x-\int_0^{\frac{\pi}{4}} \log 2 d x\end{aligned}$
In $1^{\text {st }}$ integral, $2 x= t \Rightarrow 2 d x= dt$
$\therefore \quad I=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{4} \log 2$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{4} \log 2$
$\ldots\left[\because \int_{ b }^{ b } f (x) d x=\int_{ a }^{ b } f ( t ) dt \right]$
$\begin{array}{ll}\therefore & I=\frac{1}{2} I-\frac{\pi}{4} \log 2 \\ \therefore & I=\frac{-\pi}{2} \log 2\end{array}$
View full question & answer→MCQ 572 Marks
Suppose f is such that $f (-x)=- f (x)$ for every real $x$ and $\int_0^1 f (x) d x=5$, then $\int_{-1}^0 f ( t ) dt =$
Answer(D)
Since $\int_{- a }^{ a } f (x) d x=0$, if $f (-x)=- f (x)$
$\therefore \quad \int_{-1}^1 f (x) d x=0$
$\begin{array}{l}\Rightarrow \int_{-1}^0 f (x) d x+\int_0^1 f (x) d x=0 \\ \Rightarrow \int_{-1}^0 f (x) d x=-5 \\ \Rightarrow \int_{-1}^0 f ( t ) dt =-5\end{array}$
View full question & answer→MCQ 582 Marks
If $\int_{-1}^1 f(x) d x=0$, then
- A
$f (-x)= f (x)$
- ✓
$f (-x)=- f (x)$
- C
$f (x)=2 f (x)$
- D
$f (-x)=2 f (x)$
AnswerCorrect option: B. $f (-x)=- f (x)$
(B)
$\int_{-1}^1 f (x) d x=\int_{-1}^0 f (x) d x+\int_0^1 f (x) d x$
In $1^{\text {st }}$ integral, put $x=- t \Rightarrow d x=- dt$
$\therefore \quad \int_{-1}^0 f (x) d x=-\int_1^0 f (- t ) dt$
$\begin{array}{l}=\int_0^1 f(-t) d t \\ =\int_0^1 f(-x) d x\end{array}$
$\therefore \quad \int_{-1}^1 f (x) d x=\int_0^1 f (-x) d x+\int_0^1 f (x) d x$
$=0$, if $f (-x)=- f (x)$
View full question & answer→MCQ 592 Marks
If $I _1=\int_0^{\pi / 2} f (\sin 2 x) \sin x d x$ and $I _2=\int_0^{\pi / 4} f (\cos 2 x) \cos x d x$, then $I _1 / I _2$ is equal to
- A
- B
- C
$1 / \sqrt{2}$
- ✓
$\sqrt{2}$
AnswerCorrect option: D. $\sqrt{2}$
(D)
$I _1=\int_0^{\frac{\pi}{2}} f (\sin 2 x) \sin x d x$
$=\int_0^{\frac{\pi}{4}}\left[ f (\sin 2 x) \sin x+ f \left\{\sin 2\left(\frac{\pi}{2}-x\right)\right\} \sin \left(\frac{\pi}{2}-x\right)\right] d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$=\int_0^{\frac{\pi}{4}}[ f (\sin 2 x) \sin x+ f \{\sin (\pi-2 x)\} \cos x] d x$
$\therefore \quad I _1=\int_0^{\frac{\pi}{4}}[ f (\sin 2 x) \sin x+ f (\sin 2 x) \cos x] d x \ldots$ (i)
$I _2=\int_0^{\frac{\pi}{4}} f (\cos 2 x) \cos x d x$
$=\int_0^{\frac{\pi}{4}} f \left[\cos 2\left(\frac{\pi}{4}-x\right)\right] \cdot \cos \left(\frac{\pi}{4}-x\right) d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^{\frac{\pi}{4}} f \left[\cos \left(\frac{\pi}{2}-2 x\right)\right] \cdot\left(\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x\right) d x$
$=\int_0^{\frac{\pi}{4}} f (\sin 2 x)\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right) d x$
$=\frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{4}}[ f (\sin 2 x) \cos x+ f (\sin 2 x) \sin x] d x$
$\therefore \quad I _2=\frac{1}{\sqrt{2}} I _1 \quad \ldots[$ From $( i )]$
$\therefore \quad \frac{ I _1}{ I _2}=\sqrt{2}$
View full question & answer→MCQ 602 Marks
$\int_0^\pi x f(\sin x) d x$ is equal to
- A
$\pi \int_0^\pi f(\cos x) d x$
- B
$\pi \int_0^\pi f(\sin x) d x$
- C
$\frac{\pi}{2} \int_0^{\frac{\pi}{2}} f (\sin x) d x$
- ✓
$\pi \int_0^{\frac{\pi}{2}} f (\cos x) d x$
AnswerCorrect option: D. $\pi \int_0^{\frac{\pi}{2}} f (\cos x) d x$
(D)
Let $I =\int_0^\pi x f (\sin x) d x$ ...(i)
$\therefore \quad I=\int_0^\pi(\pi-x) f(\sin x) d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I=\pi \int_0^\pi f(\sin x) d x$
$\Rightarrow 2 I =2 \pi \int_0^{\frac{\pi}{2}} f (\sin x) d x\left[\begin{array}{c}\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x, \\ \text { if } f (2 a -x)= f (x)\end{array}\right]$
$\Rightarrow I =\pi \int_0^{\frac{\pi}{2}} f (\sin x) d x$
$\Rightarrow I =\pi \int_0^{\frac{\pi}{2}} f \left(\sin \left(\frac{\pi}{2}-x\right)\right) d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\pi \int_0^{\frac{\pi}{2}} f (\cos x) d x$
View full question & answer→MCQ 612 Marks
$\int_0^\pi \frac{x d x}{4 \cos ^2 x+9 \sin ^2 x}=$
- ✓
$\frac{\pi^2}{12}$
- B
$\frac{\pi^2}{4}$
- C
$\frac{\pi^2}{6}$
- D
$\frac{\pi^2}{3}$
AnswerCorrect option: A. $\frac{\pi^2}{12}$
(A)
Let $I =\int_0^\pi \frac{x d x}{4 \cos ^2 x+9 \sin ^2 x}$ ...(i)
$\therefore \quad I =\int_0^\pi \frac{(\pi-x) d x}{4 \cos ^2(\pi-x)+9 \sin ^2(\pi-x)}$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I =\int_0^\pi \frac{(\pi-x) d x}{4 \cos ^2 x+9 \sin ^2 x}$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_0^\pi \frac{\pi d x}{4 \cos ^2 x+9 \sin ^2 x}$
$\therefore \quad I =\frac{\pi}{2} \int_0^\pi \frac{ d x}{4 \cos ^2 x+9 \sin ^2 x}$
$=2\left(\frac{\pi}{2}\right) \int_0^{\frac{\pi}{2}} \frac{d x}{4 \cos ^2 x+9 \sin ^2 x}$
$\ldots\left[\begin{array}{c}\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x \\ \text { if } f (2 a -x)= f (x)\end{array}\right]$
$\begin{array}{l}=\pi \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{4+9 \tan ^2 x} d x \\ =\frac{\pi}{9} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\frac{4}{9}+\tan ^2 x} d x\end{array}$
Put $\tan x= t \Rightarrow \sec ^2 x d x= dt$
$\therefore \quad I=\frac{\pi}{9} \int_0^{\infty} \frac{ dt }{\frac{4}{9}+ t ^2}$
$\begin{array}{l}=\frac{\pi}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 t}{2}\right]_0^{\infty} \\ =\frac{\pi}{6}\left[\frac{\pi}{2}-0\right] \\ =\frac{\pi^2}{12}\end{array}$
View full question & answer→MCQ 622 Marks
$\int_0^\pi \log \sin ^2 x d x=$
- ✓
$2 \pi \log _e\left(\frac{1}{2}\right)$
- B
$\pi \log _{ e } 2$
- C
$\frac{\pi}{2} \log _e\left(\frac{1}{2}\right)$
- D
$\pi \log _{ e }\left(\frac{1}{2}\right)$
AnswerCorrect option: A. $2 \pi \log _e\left(\frac{1}{2}\right)$
(A)
$\int_0^\pi \log \sin ^2 x d x=\int_0^\pi 2 \log \sin x d x=\int_0^{2 \frac{\pi}{2}} \log \sin x d x$
$=2 \int_0^{\frac{\pi}{2}}[\log \sin x+\log \sin (\pi-x)] d x$
$\ldots .\left[\because \int_0^{2 a } f (x) d x=\int_0^{ a }[ f (x)+ f (2 a -x)] d x\right]$
$=4 \int_0^{\frac{\pi}{2}} \log \sin x d x$
$=4 \times\left(-\frac{\pi}{2} \log 2\right)=-2 \pi \log _{ e } 2=2 \pi \log _{ e }\left(\frac{1}{2}\right)$
View full question & answer→MCQ 632 Marks
Let $I=\int_0^{100 \pi} \sqrt{(1-\cos 2 x)} d x$, then
- A
$I=0$
- ✓
$I=200 \sqrt{2}$
- C
$I=\pi \sqrt{2}$
- D
$I=100$
AnswerCorrect option: B. $I=200 \sqrt{2}$
(B)
$I=\int_0^{100 \pi} \sqrt{(1-\cos 2 x)} d x$
$\begin{array}{l}=\int_0^{100 \pi} \sqrt{2 \sin ^2 x} d x \\ =\sqrt{2} \int_0^{100 \pi} \sin x d x \\ =100 \sqrt{2} \int_0^\pi \sin x d x\end{array}$
$\ldots\left[\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x\right.$, if $\left.f (2 a -x)= f (x)\right]$
$\begin{array}{l}=100 \sqrt{2}[-\cos x]_0^\pi \\ =200 \sqrt{2}\end{array}$
View full question & answer→MCQ 642 Marks
$\int_0^{100 \pi}|\cos x| d x=$ ________.
Answer(A)
Let $I =\int_0^{100 \pi}|\cos x| d x$
$=200 \int_0^{\frac{\pi}{2}}|\cos x| d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x\right.$, if $\left.f (2 a -x)= f (x)\right]$
Since $\cos x$ is positive in the interval $\left(0, \frac{\pi}{2}\right)$
$\therefore I =200 \int_0^{\frac{\pi}{2}} \cos x d x$
$\begin{array}{l}=200[\sin x]_0^{\frac{\pi}{2}} \\ =200\end{array}$
View full question & answer→MCQ 652 Marks
If n is any integer, then $\int_0^\pi e ^{\cos ^2 x} \cos ^3(2 n +1) x d x=$
Answer(C)
Let $f (x)= e ^{\cos ^2 x} \cos ^3(2 n +1) x$
$\therefore \quad f (\pi-x)= e ^{\cos ^2(\pi-x)} \cos ^3[(2 n +1)(\pi-x)]$
$= e ^{\cos ^2 x} \cos ^3[(2 n +1) \pi-(2 n +1) x]$
$=- e ^{\cos ^2 x} \cos ^3(2 n +1) x=- f (x)$
Since $\int_0^{2 a } f (x) d x=0$, if $f (2 a -x)=- f (x)$
$\therefore \quad \int_0^\pi e ^{\cos ^2 x} \cos ^3(2 n +1) x d x=0$
View full question & answer→MCQ 662 Marks
The value of $\int_0^{2 x} \cos ^{99} x d x$ is
Answer(D)
$\int_0^{2 \pi} \cos ^{99} x d x=2 \int_0^\pi \cos ^{99} x d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x\right.$, if $\left.f (2 a -x)= f (x)\right]$
Let $I _1=\int_0^\pi \cos ^{99} x d x$
$\Rightarrow I _1=-\int_0^\pi \cos ^{99} x d x \ldots .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\Rightarrow I _1=- I _1 \Rightarrow 2 I _1=0 \Rightarrow I _1=0$
$\therefore \quad \int_0^{2 \pi} \cos ^{99} x d x=2(0)=0$
View full question & answer→MCQ 672 Marks
If $\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x$, then
- A
$f (2 a -x)=- f (x)$
- ✓
$f (2 a -x)= f (x)$
- C
$f(a-x)=-f(x)$
- D
$f ( a -x)= f (x)$
AnswerCorrect option: B. $f (2 a -x)= f (x)$
(B)
$\int_0^{2 a } f (x) d x=\int_0^{ a } f (x) d x+\int_{ a }^{2 a } f (x) d x$
Let $I _1=\int_{ a }^{2 a } f (x) d x$
Put $x=2 a - t \Rightarrow d x=- dt$
$\therefore \quad I_1=-\int_a^0 f(2 a-t) d t$
$=\int_0^a f(2 a-t) d t=\int_0^a f(2 a-x) d x$
$\therefore \quad \int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x$
$=2 \int_0^{ a } f (x) d x, \quad$ if $f (2 a -x)= f (x)$
View full question & answer→MCQ 682 Marks
If $f (x)=\frac{ e ^x}{1+ e ^x}, I _1=\int_{ f (- a )}^{ f ( a )} x g(x(1-x)) d x$ and $I _2=\int_{ f (- a )}^{ f ( a )} g (x(1-x)) d x$, then the value of $\frac{ I _2}{ I _1}$ is
Answer(D)
$f (x)=\frac{ e ^x}{1+ e ^x}$
$\therefore \quad f(a)+f(-a)=\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}$
$\Rightarrow f ( a )+ f (- a )=\frac{ e ^{ a }}{1+ e ^{ a }}+\frac{1}{1+ e ^{ a }}=1$
Using $\int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x$, we have
$I _1=\int_{ f (- a )}^{ f ( a )}(1-x) g ((1-x) x) d x$
$\ldots[\because f(a)+f(-a)=1]$
$\Rightarrow I _1=\int_{ f (- a )}^{ f ( a )} g ((1-x) x) d x-\int_{ f (- a )}^{ f ( a )} x g((1-x) x) d x$
$\Rightarrow I _1= I _2- I _1 \Rightarrow 2 I _1= I _2 \Rightarrow \frac{ I _2}{ I _1}=2$
View full question & answer→MCQ 692 Marks
Let $I _1=\int_{ a }^{\pi- a } x f (\sin x) d x, I _2=\int_{ a }^{\pi- a } f (\sin x) d x$, then $I _2$ is equal to
- A
$\frac{\pi}{2} I _1$
- B
$\pi I _1$
- ✓
$\frac{2}{\pi} I _1$
- D
$2 I _1$
AnswerCorrect option: C. $\frac{2}{\pi} I _1$
(C)
$I _1=\int_{ a }^{\pi- a } x f (\sin x) d x$
$=\int_{ a }^{\pi- a }(\pi-x) f (\sin (\pi-x)) d x$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$=\int_{ a }^{\pi- a }(\pi-x) f (\sin x) d x$
$\therefore \quad I _1=\int_{ a }^{\pi- a } \pi f (\sin x) d x- I _1$
$\Rightarrow 2 I _1=\pi I _2 \Rightarrow I _2=\frac{2}{\pi} I _1$
View full question & answer→MCQ 702 Marks
If $f(x)=f(\pi+e-x)$ and $\int_e^\pi f(x) d x=\frac{2}{e+\pi}$, then $\int_e^\pi x f(x) d x$ is equal to
- A
$\pi- e$
- B
$\frac{\pi+ e }{2}$
- ✓
- D
$\frac{\pi- e }{2}$
Answer(C)
Let $I =\int_{ e }^\pi x f (x) d x$
$=\int_{ e }^\pi( e +\pi-x) f ( e +\pi-x) d x$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$=\int_{ e }^\pi( e +\pi-x) f (x) d x$
$\ldots[\because f (x)= f (\pi+ e -x)$ (given) $]$
$\therefore \quad I=\int_e^\pi(e+\pi) f(x) d x-I$
$\Rightarrow 2 I =( e +\pi) \int_{ e }^\pi f (x) d x \Rightarrow 2 I =( e +\pi) \cdot \frac{2}{ e +\pi}$
$\Rightarrow I =1$
View full question & answer→MCQ 712 Marks
If $f (x)= f (2-x)$, then $\int_{0.5}^{1.5} x f(x) d x$ equals
AnswerCorrect option: B. $\int_{0.5}^{1.5} f(x) d x$
(B)
Let $I =\int_{0.5}^{1.5} x f (x) d x=\int_{0.5}^{1.5}(2-x) f (2-x) d x$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$=\int_{0.5}^{1.5}(2-x) f (x) d x$
$\ldots[\because f (x)= f (2-x)$ (given) $]$
$\therefore \quad I =2 \int_{0.5}^{1.5} f (x) d x- I$
$\Rightarrow 2 I =2 \int_{0.5}^{1.5} f (x) d x \Rightarrow I =\int_{0.5}^{1.5} f (x) d x$
View full question & answer→MCQ 722 Marks
If $f ( a + b -x)= f (x)$, then $\int_a^b x f(x) d x$ is equal to
- A
$\frac{ a + b }{2} \int_{ a }^{ b } f ( b -x) dx$
- ✓
$\frac{a+b}{2} \int_a^b f(x) d x$
- C
$\frac{b-a}{2} \int_a^b f(x) d x$
- D
$(a+b) \int_a^b f(x) d x$
AnswerCorrect option: B. $\frac{a+b}{2} \int_a^b f(x) d x$
(B)
Let $I =\int_{ a }^{ b } x f (x) d x$
$=\int_{ a }^{ b }( a + b -x) f ( a + b -x) d x$
$\therefore \quad I =\int_{ a }^{ b }( a + b ) f (x) d x-\int_{ a }^{ b } x f (x) d x$
$\ldots[\because f ( a + b -x)= f (x) \quad \ldots($ given $)]$
$\Rightarrow 2 I =( a + b ) \int_{ a }^{ b } f (x) d x \Rightarrow I =\frac{ a + b }{2} \int_{ a }^{ b } f (x) d x$
View full question & answer→MCQ 732 Marks
The value of $\int_{\pi / 4}^{3 \pi / 4} \frac{\phi}{1+\sin \phi} d \phi$ is
AnswerCorrect option: A. $\pi \tan \frac{\pi}{8}$
(A)
Let $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\phi}{1+\sin \phi} d \phi$ ...(i)
$=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi-\phi}{1+\sin (\pi-\phi)} d \phi$
$\ldots \left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi-\phi}{1+\sin \phi} d \phi$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi}{1+\sin (\pi-\phi)} d \phi$
On solving, we get
$2 I =2 \pi(\sqrt{2}-1)$
$\therefore \quad I=\pi(\sqrt{2}-1)=\pi \tan \frac{\pi}{8}$
View full question & answer→MCQ 742 Marks
The integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to
Answer(C)
Let $I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ ...(i)
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos (\pi-x)}$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1-\cos x}$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{2}{1-\cos ^2 x} d x$
$\therefore \quad I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \operatorname{cosec}^2 x d x$
$=-[\cot x]_{\pi / 4}^{3 \pi / 4}=2$
View full question & answer→MCQ 752 Marks
The value of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log 6-x^2\right)} d x$ is
- ✓
$\frac{1}{4} \log \frac{3}{2}$
- B
$\frac{1}{2} \log \frac{3}{2}$
- C
$\log \frac{3}{2}$
- D
$\frac{1}{6} \log \frac{3}{2}$
AnswerCorrect option: A. $\frac{1}{4} \log \frac{3}{2}$
(A)
Let $I =\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log 6-x^2\right)} d x$
Put $x^2= t \Rightarrow 2 x d x= dt$
$\therefore \quad I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t}{\sin t+\sin (\log 6-t)} d t$ ...(i)
$\therefore \quad I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t$ ...(ii)
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\frac{1}{2} \int_{\log 2}^{\log 3} dt =\frac{1}{2}(\log 3-\log 2)=\frac{1}{2} \log \left(\frac{3}{2}\right)$
$\Rightarrow I =\frac{1}{4} \log \left(\frac{3}{2}\right)$
View full question & answer→MCQ 762 Marks
The integral $\int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x$ is equal to
Answer(C)
$\text { Let } I=\int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x$ ...(i)
$\therefore \quad I =\int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2+\log x^2} d x$ ... (ii)
$\ldots \left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_2^4 1 d x=[x]_2^4=4-2=2$
$\therefore \quad I =1$
Alternate Method:
Here, $f (x)=\log x^2, a =2$ and $b =4$
$\int_{ a }^{ b } \frac{ f (x)}{ f (x)+ f ( a + b -x)} d x=\frac{1}{2}(b- a )$
$\therefore \int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x=\frac{1}{2}(4-2)=1$
View full question & answer→MCQ 772 Marks
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{e^{\sin x}+1}$ is equal to
- A
$0$
- B
- C
$-\frac{\pi}{2}$
- ✓
$\frac{\pi}{2}$
AnswerCorrect option: D. $\frac{\pi}{2}$
(D)
Let $I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{ e ^{\sin x}+1}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{ e ^{\sin x}\left(1+ e ^{-\sin x}\right)}$
$\therefore I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{ e ^{-\sin x}}{1+ e ^{-\sin x}} d x$ ...(i)
Also, $I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{ e ^{\sin x}+1} d x$
$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{ e ^{\sin (-x)}+1} d x$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore \quad I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{e^{-\sin x}+1} d x$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d x$
$\Rightarrow 2 I =[x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \Rightarrow I =\frac{\pi}{2}$
View full question & answer→MCQ 782 Marks
$\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}$ is
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{12}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: C. $\frac{\pi}{12}$
(C)
Let $I =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx$ ...(i)
$\therefore \quad I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ ...(ii)
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x$
$\therefore \quad I =\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$
Alternate Method:
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Here $f (x)=\sqrt{\sin x}, a =\frac{\pi}{6}$ and $b =\frac{\pi}{3}$
$\int_{ a }^{ b } \frac{ f (x)}{ f (x)+ f ( a + b -x)} d x=\frac{1}{2}(b- a )$
$\therefore \quad \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}=\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$
View full question & answer→MCQ 792 Marks
The value of $I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$ is
Answer(B)
$I =\int_{\pi / 2}^{5 \pi / 2} \frac{ e ^{\tan ^{-1}(\sin x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$
$=\int_{\pi / 2}^0 \frac{ e ^{\tan ^{-1}(\sin x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$
$+\int_0^{5 \pi / 2} \frac{ e ^{\tan ^{-1}(\sin x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$
$\therefore \quad I =\int_0^{5 \pi / 2} \frac{ e ^{\tan ^{-1}(\sin x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$ $-\int_0^{\pi / 2} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$ ...(i)
$=\int_0^{5 \pi / 2} \frac{ e ^{\tan ^{-1} \sin \left(\frac{5 \pi}{2}-x\right)}}{ e ^{\tan ^{-1} \sin \left(\frac{5 \pi}{2}-x\right)}+ e ^{\tan ^{-1} \cos \left(\frac{5 \pi}{2}-x\right)}} d x$ $-\int_0^{\pi / 2} \frac{ e ^{\tan ^{-1} \sin \left(\frac{\pi}{2}-x\right)}}{ e ^{\tan ^{-1} \sin \left(\frac{\pi}{2}-x\right)}+ e ^{\tan ^{-1} \cos \left(\frac{\pi}{2}-x\right)}} d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I =\int_0^{5 \pi / 2} \frac{ e ^{\tan ^{-1}(\cos x)}}{ e ^{\tan ^{-1}(\cos x)}+ e ^{\tan -1(\sin x)}} d x$ $-\int_0^{\pi / 2} \frac{e^{\tan ^{-1}(\cos x)}}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x \ldots$(ii)
Adding (i) and (ii), we get
$2 I =\int_0^{5 \pi / 2} \frac{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$ $-\int_0^{\pi / 2} \frac{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}}{ e ^{\tan ^{-1}(\sin x)}+ e ^{\tan ^{-1}(\cos x)}} d x$
$\begin{array}{l}\Rightarrow 2 I=\frac{5 \pi}{2}-\frac{\pi}{2}=2 \pi \\ \Rightarrow I=\pi\end{array}$
View full question & answer→MCQ 802 Marks
$\int_0^\pi[\cot x] d x$, [.] denotes the greatest integer function, is equal to
- A
$\frac{\pi}{2}$
- B
- C
- ✓
$-\frac{\pi}{2}$
AnswerCorrect option: D. $-\frac{\pi}{2}$
(D)
Let $I =\int_0^\pi[\cot x] d x$ ...(i)
$\Rightarrow I =\int_0^\pi[\cot (\pi-x)] d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\Rightarrow I=\int_0^\pi[-\cot x] d x$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_0^\pi\{[\cot x]+[-\cot x]\} d x$
$\Rightarrow 2 I =\int_0^\pi-1 d x \ldots .[\because[x]+[-x]=-1$, if $x \notin Z ]$
$\Rightarrow 2 I =-\pi$
$\Rightarrow I=-\frac{\pi}{2}$
View full question & answer→MCQ 812 Marks
$\int_0^x \frac{x \tan x}{\sec x+\tan x} d x=$
AnswerCorrect option: D. $\pi\left(\frac{\pi}{2}-1\right)$
(D)
Let $I =\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ ...(i)
$\therefore \quad I=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\tan x} d x$ ...(ii)
$\ldots .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \frac{\tan x}{\sec x+\tan x} d x$
$\therefore \quad I =\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$
$=\frac{\pi}{2}\left[\int_0^\pi 1 d x-\int_0^\pi \frac{ d x}{1+\sin x}\right]$
On solving, we get
$I=\frac{\pi}{2}(\pi-2)=\pi\left(\frac{\pi}{2}-1\right)$
View full question & answer→MCQ 822 Marks
$\int_0^x \frac{x \tan x}{\sec x+\cos x} d x=$
- ✓
$\frac{\pi^2}{4}$
- B
$\frac{\pi^2}{2}$
- C
$\frac{3 \pi^2}{2}$
- D
$\frac{\pi^2}{3}$
AnswerCorrect option: A. $\frac{\pi^2}{4}$
(A)
$\text { Let } I=\int_0^\pi \frac{x \tan x}{\sec x+\cos x} d x$ ...(i)
$\therefore \quad I=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\cos x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \frac{\tan x}{\sec x+\cos x} d x$
$\Rightarrow I =\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$
Put $\cos x= t \Rightarrow \sin x d x=- dt$
$\therefore \quad I=-\frac{\pi}{2} \int_1^{-1} \frac{d t}{1+t^2}=-\frac{\pi}{2}\left[\tan ^{-1} t\right]_1^{-1}$
$=\left(-\frac{\pi}{2}\right)\left(-\frac{\pi}{2}\right)=\frac{\pi^2}{4}$
View full question & answer→MCQ 832 Marks
$\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x d x}{\sin ^4 x+\cos ^4 x}=$
- A
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{8}$
AnswerCorrect option: D. $\frac{\pi}{8}$
(D)
Let $I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x}{\sin ^4 x+\cos ^4 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sin ^3\left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$\therefore \quad 2 I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x+\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos ^4 x\left(1+\tan ^4 x\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\tan x \sec ^2 x}{\left(1+\tan ^4 x\right)} d x$
Put $\tan ^2 x= t$
$\therefore \quad \tan x \sec ^2 x d x=\frac{ dt }{2}$
When $x=0, t =0$ and when $x=\frac{\pi}{2}, t =\infty$
$\therefore \quad 2 I =\frac{1}{2} \int_0^{\infty} \frac{ dt }{1+ t ^2}$
$\therefore \quad I=\frac{1}{4}\left[\tan ^{-1} t\right]_0^{\infty}$
$\begin{array}{l}=\frac{1}{4}\left(\frac{\pi}{2}-0\right) \\ =\frac{\pi}{8}\end{array}$
View full question & answer→MCQ 842 Marks
$\int_0^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x=$
- A
$0$
- B
$\frac{\pi}{8}$
- C
$\frac{\pi^2}{8}$
- ✓
$\frac{\pi^2}{16}$
AnswerCorrect option: D. $\frac{\pi^2}{16}$
(D)
Let $I =\int_0^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$ ...(i)
$\therefore \quad I=\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\sin ^4 x+\cos ^4 x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\frac{\pi}{2} \int_0^{\pi / 2} \frac{\cos x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$\Rightarrow I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x$
Put $\tan ^2 x= t \Rightarrow \tan x \sec ^2 x d x=\frac{ dt }{2}$
$\therefore \quad I=\frac{\pi}{8} \int_0^{\infty} \frac{ dt }{1+ t ^2}=\frac{\pi}{8}\left[\tan ^{-1} t \right]_0^{\infty}=\frac{\pi^2}{16}$
View full question & answer→MCQ 852 Marks
The value of $\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x$ is
Answer(B)
$\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x$
$=\int_0^1 \tan ^{-1}\left(\frac{x+(x-1)}{1-x(x-1)}\right) d x$
$=\int_0^1\left(\tan ^{-1} x+\tan ^{-1}(x-1)\right) d x$
$=\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1}(x-1) d x$
$=\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1}(1-x-1) d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^1 \tan ^{-1} x d x-\int_0^1 \tan ^{-1} x d x=0$
View full question & answer→MCQ 862 Marks
The value of $\int_0^1 \frac{\log (1+x)}{1+x^2} d x$ is
- A
$\frac{\pi}{2} \log 2$
- B
$\pi \log 2$
- C
$\log 2$
- ✓
$\frac{\pi}{8} \log 2$
AnswerCorrect option: D. $\frac{\pi}{8} \log 2$
(D)
Let $I =\int_0^1 \frac{\log (1+x)}{1+x^2} d x$
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\therefore \quad I=\int_0^{\pi / 4} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \cdot \sec ^2 \theta d \theta$
$=\int_0^{\pi / 4} \frac{\log (1+\tan \theta)}{\sec ^2 \theta} \cdot \sec ^2 \theta d \theta$
$\therefore \quad I=\int_0^{\pi / 4} \log (1+\tan \theta) d \theta$
$\int_0^{\pi / 4} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$
$\Rightarrow I=\frac{\pi}{8} \log 2 $
View full question & answer→MCQ 872 Marks
$\int_0^{\pi / 4} \log (1+\tan \theta) d \theta=$
AnswerCorrect option: C. $\frac{\pi}{8} \log 2$
(C)
Let $I=\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta$
$=\int_0^{\frac{\pi}{4}} \log 2 d \theta-\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\therefore \quad 2 I=\int_0^{\frac{\pi}{4}} \log 2 d \theta \Rightarrow I=\frac{\log 2}{2}[\theta]_0^{\pi / 4}=\frac{\pi}{8} \log 2$
View full question & answer→MCQ 882 Marks
$\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x$
- A
$\frac{\pi-1}{2}$
- ✓
$\frac{\pi-1}{4}$
- C
$\frac{1+\pi}{4}$
- D
$\frac{\pi-3}{4}$
AnswerCorrect option: B. $\frac{\pi-1}{4}$
(B)
Let $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x$ …(i)
$I=\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos x+\sin x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x+\cos ^3 x}{\sin x+\cos x} d x$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}}\left(\sin ^2 x-\sin x \cos x+\cos ^2 x\right) d x \\ =\int_0^{\frac{\pi}{2}}(1-\sin x \cos x) d x\end{array}$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}} 1 d x-\int_0^{\frac{\pi}{2}} \sin x \cos x d x \\ =[x]_0^{\pi / 2}-\left[\frac{\sin ^2 x}{2}\right]_0^{\pi / 2}=\frac{\pi}{2}-\frac{1}{2}\end{array}$
$\therefore \quad 2 I=\frac{\pi-1}{2} \Rightarrow I=\frac{\pi-1}{4}$
View full question & answer→MCQ 892 Marks
$\int_0^\pi \frac{x}{1+\sin x} d x$ is equal to
- A
$\frac{\pi}{2}$
- ✓
$\pi$
- C
$\frac{\pi}{2} \log 2$
- D
$\pi \log 2$
Answer(B)
Let $I=\int_0^\pi \frac{x}{1+\sin x} d x$ ...(i)
$\therefore \quad I=\int_0^\pi \frac{\pi-x}{1+\sin x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_0^\pi \frac{\pi}{1+\sin x} d x$
$\begin{array}{l}=\pi \int_0^\pi \frac{1-\sin x}{(1+\sin x)(1-\sin x)} d x \\ =\pi \int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x=\pi \int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x\end{array}$
$=\pi[\tan x-\sec x]_0^\pi$
$\therefore \quad 2 I =\pi[0-(-1)-(0-1)]=2 \pi$
$\Rightarrow I =\pi$
View full question & answer→MCQ 902 Marks
$\int_0^\pi x \sin ^3 x d x=$
- A
$\frac{4 \pi}{3}$
- ✓
$\frac{2 \pi}{3}$
- C
$0$
- D
$\frac{\pi}{4}$
AnswerCorrect option: B. $\frac{2 \pi}{3}$
(B)
Let $I =\int_0^\pi x \sin ^3 x d x$ ...(i)
$=\int_0^\pi(\pi-x) \sin ^3 x d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I=\pi \int_0^\pi \sin ^3 x d x=\frac{\pi}{4} \int_0^\pi(3 \sin x-\sin 3 x) d x$
$\begin{array}{l}=\frac{\pi}{4}\left[-3 \cos x+\frac{\cos 3 x}{3}\right]_0^\pi \\ =\frac{\pi}{4}\left[3-\frac{1}{3}+3-\frac{1}{3}\right]=\frac{4 \pi}{3}\end{array}$
$\therefore \quad I=\frac{2 \pi}{3}$
View full question & answer→MCQ 912 Marks
Let $f$ be a non-constant continuous function for all $x \geq 0$. Let $f$ satisfy the relation $f (x) f ( a -x)=1$ for some $a \in R ^{+}$. Then $I=\int_0^a \frac{d x}{1+f(x)}$ is equal to
- A
- B
$\frac{a}{4}$
- ✓
$\frac{a}{2}$
- D
$f(a)$
AnswerCorrect option: C. $\frac{a}{2}$
(C)
$I=\int_0^a \frac{d x}{1+f(x)}$ ...(i)
$=\int_0^{ a } \frac{ d x}{1+ f ( a -x)} \quad \ldots\left[\int_0^{ a } f (x)=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^{ a } \frac{ d x}{1+\frac{1}{ f (x)}} \quad \ldots[\because f (x) f ( a -x)=1]$
$\therefore \quad I=\int_0^a \frac{f(x)}{1+f(x)} d x$ ...(ii)
Adding (i) and (ii), we get
$\begin{array}{l}2 I =\int_0^{ a } d x=[x]_0^{ a } \\ \Rightarrow I =\frac{ a }{2}\end{array}$
View full question & answer→MCQ 922 Marks
$\int_0^{\frac{\alpha}{3}} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$
- A
$\frac{2 \alpha}{3}$
- B
$\frac{\alpha}{2}$
- C
$\frac{\alpha}{3}$
- ✓
$\frac{\alpha}{6}$
AnswerCorrect option: D. $\frac{\alpha}{6}$
(D)
Let $I=\int_0^{\frac{\alpha}{3}} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x$
$\Rightarrow I =\int_0^{\frac{\alpha}{3}} \frac{ f (x)}{ f (x)+ f \left(\frac{\alpha}{3}-x\right)} d x$
$\int_0^{ a } \frac{ f (x)}{ f (x)+ f ( a -x)} d x=\frac{ a }{2}$
$=\frac{\frac{\alpha}{3}}{2} $
$=\frac{\alpha}{6}$
View full question & answer→MCQ 932 Marks
The value of $\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is
Answer(C)
Let $I =\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$
$=\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$
$\ldots .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I=-\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x=-I$
$\Rightarrow 2 I =0 \Rightarrow I =0$
View full question & answer→MCQ 942 Marks
The value of $\int_0^\pi e ^{\cos ^2 x} \cos ^5 3 x d x$ is
Answer(C)
Let $I =\int_0^\pi e ^{\cos ^2 x} \cos ^5 3 x d x$
$=\int_0^\pi e ^{\cos ^2(\pi-x)} \cos ^5 3(\pi-x) d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I =-\int_0^\pi e ^{\cos ^2 x} \cos ^5 3 x d x=- I$
$\Rightarrow 2 I =0 \Rightarrow I =0$
View full question & answer→MCQ 952 Marks
$\int_0^{\pi / 2} \frac{d \theta}{1+\tan \theta}=$
- A
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
Let $I=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\tan \theta}$ …(i)
$=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\tan \left(\frac{\pi}{2}-\theta\right)}$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I=\int_0^{\frac{\pi}{2}} \frac{d \theta}{1+\cot \theta}$ ...(ii)
Adding (i) and (ii), we get
$2 I=\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan \theta}+\frac{1}{1+\cot \theta}\right) d \theta$
$=\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan \theta}+\frac{\tan \theta}{\tan \theta+1}\right) d \theta$
$=\int_0^{\frac{\pi}{2}} d \theta=[\theta]_0^{\pi / 2}$
$\therefore \quad 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$
Alternate Method:
$\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^{ n } x}=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\cot ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\pi / 2} \frac{d \theta}{1+\tan \theta}=\frac{\pi}{4}$
View full question & answer→MCQ 962 Marks
The value of $\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is
- A
$0$
- B
- C
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
Let $I=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^3 x}$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i)
$\therefore \quad I=\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos ^3 x+\sin ^3 x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_0^{\frac{\pi}{2}} d x=[x]_0^{\pi / 2}$
$\therefore \quad 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$
Alternate Method:
$\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^{ n } x}=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\cot ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}=\frac{\pi}{4}$
View full question & answer→MCQ 972 Marks
The value of $\int_0^1 \frac{d x}{x+\sqrt{1-x^2}}$ is
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{2}$
- C
$\frac{1}{2}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
$\int_0^1 \frac{d x}{x+\sqrt{1-x^2}}=\int_0^{\pi / 2} \frac{\cos \theta d \theta}{\sin \theta+\cos \theta}$
$\ldots . .[$ Put $x=\sin \theta \Rightarrow d x=\cos \theta d \theta]$
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\frac{\pi}{4}$
$=\frac{\pi}{4} $
View full question & answer→MCQ 982 Marks
$\int_0^{\pi / 2} \frac{\sin ^{\frac{3}{2}} x d x}{\cos ^{\frac{3}{2}} x+\sin ^{\frac{3}{2}} x}=$
- A
$0$
- B
$\pi$
- C
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
(D)
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\cos ^{\frac{3}{2}} x+\sin ^{\frac{3}{2}} x} d x=\frac{\pi}{4} $
View full question & answer→MCQ 992 Marks
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{1000} x d x}{\sin ^{1000} x+\cos ^{1000} x}$ is equal to
- A
- ✓
$\frac{\pi}{4}$
- C
- D
$\frac{\pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{4}$
(B)
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\int_0^{\frac{\pi}{2}} \frac{\cos ^{ n } x}{\sin ^{ n } x+\cos ^{ n } x} d x=\frac{\pi}{4}$
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} d x=\frac{\pi}{4}$
View full question & answer→MCQ 1002 Marks
The value of $\int_0^{\frac{\pi}{2}} \frac{ e ^{x^2}}{ e ^{x^2}+ e ^{\left(\frac{\pi}{2}-x\right)^2}} d x$ is
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$e^{\frac{\pi^2}{16}}$
- D
$e^{\frac{\pi^2}{4}}$
AnswerCorrect option: A. $\frac{\pi}{4}$
(A)
Here, $f (x)= e ^{x^2}$ and $a =\frac{\pi}{2}$
$\int_0^{ a } \frac{ f (x)}{ f (x)+ f ( a -x)} d x=\frac{ a }{2}$
$\int_0^{\frac{\pi}{2}} \frac{ e ^{x^2}}{ e ^{x^2}+ e ^{\left(\frac{\pi}{2}-x\right)^2}} d x=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}$
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