1 Marks Question

Question 11 Mark

Evaluate $\int_0^1 \frac{d x}{1+x^2}$.

Answer

$\int_0^1 \frac{d x}{1+x^2}=\left(\tan ^{-1} x\right)_0^1$
$=\tan ^{-1} 1-\tan ^{-1} 0=\frac{\pi}{4}-0=\frac{\pi}{4}$

Question 21 Mark

Evaluate $\int \frac{\cos ^2 x}{1+\sin x} d x$

Answer

Let
$I=\int \frac{\cos ^2 x}{1+\sin x} d x$
$I=\int \frac{1-\sin ^2 x}{1+\sin x} d x$
$=\int \frac{(1-\sin x)(1+\sin x)}{1+\sin x} d x$
$I=\int(1-\sin x) d x$
$=x+\cos x+\text { C }$

View full question & answer→

Question 31 Mark

Evaluate $\int \sqrt{1-\sin 2 x}\ d x$ :

Answer

$\int \sqrt{1-\sin 2 x}\ d x$
$=\int \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}\ d x$
$=\int \sqrt{(\sin x-\cos x)^2} d x$
$=\int(\sin x-\cos x) d x$
$=\int \sin x\ d x-\int \cos x\ d x$
$=-\cos x-\sin x+C$

View full question & answer→

Question 41 Mark

Evaluate $\int_0^\pi \sin ^2 x d x$.

Answer

Let
$
\begin{aligned}
I & =\int_0^\pi \sin ^2 x d x=\int_0^\pi\left(\frac{1-\cos 2 x}{2}\right) d x \\
I & =\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)_0^\pi \\
& =\frac{1}{2}\left(\left(\pi-\frac{1}{2} \sin 2 \pi\right)-(0-0)\right) \\
& =\frac{1}{2}(\pi-0)=\frac{\pi}{2} \text { }
\end{aligned}
$

View full question & answer→

Question 51 Mark

Find the value of $\int_{-4}^4|x| d x$.

Answer

Let $ I =\int_{-4}^4|x| d x=\int_{-4}^0|x| d x+\int_0^4|x| d x$
$ \because|x|=\left\{\begin{array}{l} x, x>0 \\ -x, x<0\end{array}\right.$
$\therefore =\int_{-4}^0 - x\ d x+\int_0^4 x\ d x$
$ =\left(\frac{-x^2}{2}\right)_{-4}^0+\left(\frac{x^2}{2}\right)_0^4$
$ =\frac{-1}{2}(0-16)+\frac{1}{2}(16-0)$
$=8+8$
$=16$

View full question & answer→

Question 61 Mark

Solve: $\int \cos ^{-1}(\sin x) d x$

Answer

$
\begin{aligned}
I & =\int \cos ^{-1}(\sin x) d x \\
& =\int \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-x\right)\right) d x \\
& =\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} \int d x-\int x d x \\
& =\frac{\pi}{2} x-\frac{x^2}{2}+C \quad \text { Ans. }
\end{aligned}
$

View full question & answer→

Question 71 Mark

Find the value of $\int \log _a x d x$.

Answer

Find the value of $\int \log _a x d x$. Let $\quad I =\int \log _a x d x$
On changing the base
$
I=\int \log _e x \cdot \log _a e d x
$
$
=\log _a e \int_{II} 1 \cdot \log _e x d x
$
Taking 1 as the second function, integrating by parts :
$
\begin{aligned}
I & =\log _a e\left[\log _e x \cdot x-\int \frac{1}{x} \cdot x d x\right] \\
& =\log _a e\left(x \log _e x-x\right)+C \\
& =x \log _a e\left(\log _e x-1\right)+C \text {}
\end{aligned}
$

View full question & answer→

Question 81 Mark

Write the value of $\int x \operatorname{cosec}^2 x d x$ :

View full question & answer→

Question 91 Mark

Find the value of $\int \frac{x d x}{\left(1-x^2\right)^{3 / 2}}$ :

Answer

$\int \frac{x d x}{\left(1-x^2\right)^{3 / 2}}$
Let
$1-x^2=t $
$\Rightarrow-2 x\ d x=d t$
$\therefore x \ d x=\frac{-1}{2} d t$
$\Rightarrow I =\frac{-1}{2} \int \frac{d t}{t^{3 / 2}}$
$=\frac{-1}{2} \int t^{-3 / 2} d t$
$=\frac{-1}{2} \frac{t^{-1 / 2}}{-1 / 2}+ C $
$=\frac{1}{\sqrt{1-x^2}}+ C \quad$

View full question & answer→

Question 101 Mark

Find the value of $\int \frac{e^x d x}{\sqrt{1-e^{2 x}}}$ :

Answer

Let $\quad I =\int \frac{e^x d x}{\sqrt{1-\left(e^x\right)^2}}$
Let $\quad e^x=t$
$\Rightarrow \quad e^x d x=d t$
$I =\int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1} t+ C =\sin ^{-1} e^x+ C$

View full question & answer→

Question 111 Mark

Integrate the following function :
$
\cos ^3 x \cdot e^{\log _e \sin x}
$

View full question & answer→

Question 121 Mark

Find the value of $\int(1+x)\left(1+x^2\right)(1-x) d x$ :

Answer

$\int(1+x)\left(1+x^2\right)(1-x) d x$
$\Rightarrow \int(1+x)(1-x)\left(1+x^2\right) d x$
$\Rightarrow \int\left(1-x^2\right)\left(1+x^2\right) d x$
$\Rightarrow \int\left(1-x^4\right) d x=\int d x-\int x^4 d x$
$\Rightarrow x-\frac{x^5}{5}+C$

View full question & answer→

Question 131 Mark

Evaluate $\int e^{3 \log _e x} d x$ :

Answer

Let:
$
\begin{aligned}
I & =\int e^{3 \log _e x} d x \\
& =\int e^{\log _e x^3} d x=\int x^3 d x=\frac{1}{4} x^4+C
\end{aligned}
$

View full question & answer→

Question 141 Mark

Evaluate $\int \sqrt[3]{x^4} d x$.

Answer

Let
$
I=\int \sqrt(3){x^4} d x
$
$\Rightarrow \quad \int\left(x^4\right)^{\frac{1}{3}} d x=\int x^{\frac{4}{3}} d x$
$\Rightarrow \quad \frac{x^{\frac{4}{3}+1}}{\frac{4}{3}+1}=\frac{3}{7} x^{\frac{7}{3}}+ C$

View full question & answer→

Question 151 Mark

Evalaute:
$
\int_0^2 \sqrt{4-x^2} d x
$

Answer

Let
$
\begin{aligned}
I & =\int_0^2 \sqrt{4-x^2} d x \\
& =\int_0^2 \sqrt{(2)^2-x^2} d x
\end{aligned}
$
We know that
$
\int \sqrt{a^2-x^2} d x=\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1}\left(\frac{x}{a}\right)+C
$
So
$
I=\left(\frac{1}{2} x \sqrt{4-x^2}+\frac{1}{2} \times 4 \sin ^{-1}\left(\frac{x}{2}\right)\right)_0^2
$
$\begin{array}{c}=\left[0+2 \sin ^{-1}(1)-0\right]=2 \sin ^{-1} 1 \\ =2 \times \frac{\pi}{2}=\pi \text { }\end{array}$

View full question & answer→

Question 161 Mark

Evaluate $\int_0^a \frac{d x}{\sqrt{a^2-x^2}}$.

Answer

Let
$
\text { } \begin{aligned}
I & =\int_0^a \frac{d x}{\sqrt{a^2-x^2}} \\
=\left(\sin ^{-1} \frac{x}{a}\right)_0^a & =\sin ^{-1} \frac{a}{a}-\sin ^{-1}(0) \\
& =\sin ^{-1} 1-\sin ^{-1} 0 \\
& =\frac{\pi}{2}-0=\frac{\pi}{2} \text { Ans. }
\end{aligned}
$

View full question & answer→

Question 171 Mark

Find the value of $\int \frac{1}{x} d x$ :

Answer

$\int \frac{1}{x} d x=\log x+C$

View full question & answer→

Question 181 Mark

Find the value of $\int 2^x d x$ :

Answer

$\int 2^x d x=\frac{2^x}{\log 2}+C$
Since we know that,
$
\int a^x d x=\frac{a^x}{\log _e a}+C
$

View full question & answer→

Question 191 Mark

Evaluate $\int_0^1 \frac{\left(x^2-x\right)}{\sqrt{x}} d x$.

Answer

Let
$
\begin{aligned}
I & =\int_0^1 \frac{\left(x^2-x\right)}{\sqrt{x}} d x=\int_0^1\left(x^{3 / 2}-x^{\frac{1}{2}}\right) d x \\
I & =\left[\frac{x^{5 / 2}}{5 / 2}-\frac{x^{3 / 2}}{3 / 2}\right]_0^1=\left[\frac{2}{5}(1-0)-\frac{2}{3}(1-0)\right] \\
& =\frac{2}{5}-\frac{2}{3}=\frac{6-10}{15}=\frac{-4}{15} \text { }
\end{aligned}
$

View full question & answer→

Question 201 Mark

Find the value of $\int \frac{\sin x^{1 / 3}}{x^{2 / 3}} d x$ :

Answer

$\int \frac{\sin x^{1 / 3}}{x^{2 / 3}} d x$
Putting
$x^{1 / 3} =t$
$\frac{1}{3} x^{\frac{1}{3}-1} d x =d t$
So
$x^{-2 / 3} d x =3 d t$
$\frac{d x}{x^{2 / 3}} =3 d t$
$\Rightarrow 3 \int \sin t\ d t $
$\Rightarrow 3(-\cos t)+C$
$\Rightarrow-3 \cos t+C $
$\Rightarrow-3 \cos x^{\frac{1}{3}}+C$

View full question & answer→

Question 211 Mark

Find the value of $\int b^{2 x} d b$ :

Answer

$\int b^{2 x} d b$
Here we need to integrate with respect to $x$
$
\therefore \int b^{2 x} d b=\frac{b^{2 x+1}}{2 x+1}+C \quad\left(\because \int b^n d b=\frac{b^{n+1}}{n+1}+C\right)
$

View full question & answer→

MATHS 1 Marks Question

Take a timed test