Exercise 2.4 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.4$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$
2 \cos x-3 \sin x+4 \sec ^2 x-5 \operatorname{cosec}^2 x
$
Solution:

$\begin{aligned}
& \int 2 \cos x-3 \sin x+4 \sec ^2 x-5 \operatorname{cosec}^2 x \\
& =2 \int \cos x d x-3 \int \sin x d x+4 \int \sec ^2 x-5 \int \operatorname{cosec}^2 x d x \\
& =2 \sin x+3 \cos x+4 \tan x+5 \cot x+c
\end{aligned}$
Question 2.
$\int \sin ^3 \mathrm{x} d x$
Solution:
We know that, $\sin 3 \mathrm{x}=3 \sin \mathrm{x}-4 \sin ^3 \mathrm{x}$
$
\begin{aligned}
\Rightarrow \sin ^3 x & =\frac{3 \sin x-\sin 3 x}{4} \\
\text { So } \int \sin ^3 x d x & =\int \frac{(3 \sin x-\sin 3 x)}{4} d x \\
& =\frac{3}{4} \int \sin x d x-\frac{1}{4} \int \sin 3 x d x+c \\
& =\frac{-3}{4} \cos x-\frac{1}{4} \frac{\cos 3 x}{(-3)}+c \\
\text { (Using } \int \sin a x & \left.=\frac{-1}{a} \cos a x\right) \\
& =-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x+c
\end{aligned}
$
Question 3.
$\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}$
Solution:

$
\begin{aligned}
\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} & =\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x} \\
& =\frac{1}{\cos ^2 x}=\sec ^2 x \\
\text { So } \int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x & =\int \sec ^2 x d x=\tan x+c
\end{aligned}
$
Question 4.
$
\begin{aligned}
& \frac{1}{\sin ^2 x \cos ^2 x} \\
& {\left[\text { Hint: } \sin ^2 \mathrm{x}+\cos ^2 \mathrm{x}=1\right]}
\end{aligned}
$
Solution:
$
\begin{aligned}
\frac{1}{\sin ^2 x \cos ^2 x} & =\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}=\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x} \\
& =\sec ^2 x+\operatorname{cosec}^2 x \\
\text { So } \int \frac{1}{\sin ^2 x \cos ^2 x} d x & =\int \sec ^2 x d x+\int \operatorname{cosec}^2 x d x+c \\
& =\tan x-\cot x+c
\end{aligned}
$
Question 5.
$\sqrt{1-\sin 2} x$
Solution:
$
\begin{aligned}
\sqrt{1-\sin 2 x} & =\sqrt{1-2 \sin x \cos x} \\
& =\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x} \\
& =\sqrt{(\cos x-\sin x)^2} \\
& =\cos x-\sin x
\end{aligned}
$
$
\begin{aligned}
& \text { So } \int \sqrt{1-\sin 2} x d x=\int \cos x d x-\int \sin x+c \\
& =\sin x-(-\cos x)+c \\
& =\sin x+\cos x+c \\
&
\end{aligned}
$

$\begin{aligned}
& \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}=\sqrt{(\sin x-\cos x)^2} \\
& =\sin x-\cos x \\
& \text { So } \int \sqrt{1-\sin 2 x d x}=\int \sin x d x-\int \cos x d x+c \\
& =-\cos x-\sin x+c \\
& =-(\cos x+\sin x)+c \\
&
\end{aligned}$