Exercise 11.3 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.3
Integrate the following w.r.to $\mathrm{x}$.
Question 1 .
$
(\mathrm{x}+4)^5+\frac{5}{(2-5 x)^4}-\operatorname{cosec}^2(3 \mathrm{x}-1)
$
Solution:
$
\begin{aligned}
& \int\left[(x+4)^5+\frac{5}{(2-5 x)^4}-\operatorname{cosec}^2(3 x-1)\right] d x \\
&=\int(x+4)^5 d x+5 \int \frac{1}{(2-5 x)^4} d x-\int \operatorname{cosec}^2(3 x-1) d x \\
&=\frac{(x+4)^6}{6}+\frac{-5}{-5 \times 3(2-5 x)^3}+\frac{\cot (3 x-1)}{3}+c \\
&= \frac{(x+4)^6}{6}+\frac{1}{3(2-5 x)^3}+\frac{\cot (3 x-1)}{3}+c
\end{aligned}
$
Question 2.
$
\begin{aligned}
& 4 \cos (5-2 \mathrm{x})+9 \mathrm{e}^{3 \mathrm{x}-6}+\frac{24}{6-4 x} \\
& \text { Solution: } \\
& \int\left[4 \cos (5-2 x)+9 e^{3 x-6}+\frac{24}{6-4 x}\right] d x \\
& =4 \int \cos (5-2 x) d x+9 \int e^{3 x-6} d x+24 \int \frac{1}{6-4 x} d x \\
& =\frac{-4}{2} \cos (5-2 x)+\frac{9}{3} e^{3 x-6}+\frac{24}{-4} \log |6-4 x|+c \\
& =-2 \cos (5-2 x)+3 e^{3 x-6}-6 \log |6-4 x|+c
\end{aligned}
$
Question 3.
$
\sec ^2 \frac{x}{5}+18 \cos 2 x+10 \sec (5 x+3) \tan (5 x+3)
$

Solution:
$
\begin{aligned}
& \int\left[\sec ^2 \frac{x}{5}+18 \cos 2 x+10 \sec (5 x+3) \tan (5 x+3)\right] d x \\
& \quad=\int \sec ^2 \frac{x}{5} d x+18 \int \cos 2 x d x+10 \int \sec (5 x+3) \tan (5 x+3) d x \\
& \quad=\frac{\tan \frac{x}{5}}{1 / 5}+18\left(\frac{\sin 2 x}{2}\right)+10 \frac{\sec (5 x+3)}{5}+c \\
& \quad=5 \tan x / 5+9 \sin 2 x+2 \sec (5 x+3)+c
\end{aligned}
$
Question 4.
$
\begin{aligned}
& \frac{8}{\sqrt{1-(4 x)^2}}+\frac{27}{\sqrt{1-9 x^2}}-\frac{15}{1+25 x^2} \\
& \text { Solution: } \\
& \int\left[\frac{8}{\sqrt{1-(4 x)^2}}+\frac{27}{\sqrt{1-9 x^2}}-\frac{15}{1+25 x^2}\right] d x \\
& =8 \int \frac{1}{\sqrt{1-(4 x)^2}} d x+27 \int \frac{1}{\sqrt{1-(3 x)^2}} d x-15 \int \frac{1}{\sqrt{1+(5 x)^2}} d x \\
& =\frac{8}{4} \sin ^{-1}(4 x)+\frac{27}{3} \sin ^{-1}(3 x) \frac{-15}{5} \tan ^{-1}(5 x)+c \\
& =2 \sin ^{-1}(4 x)+9 \sin ^{-1}(3 x)-3 \tan ^{-1}(5 x)+c
\end{aligned}
$
Question 5.
$
\frac{6}{1+(3 x+2)^2}-\frac{12}{\sqrt{1-(3-4 x)^2}}
$

Solution:
$
\begin{aligned}
& \int\left[\frac{6}{1+(3 x+2)^2}-\frac{12}{\sqrt{1-(3-4 x)^2}}\right] d x \\
& =6 \int \frac{1}{1+(3 x+2)^2} d x-12 \int \frac{1}{\sqrt{1-(3-4 x)^2}} d x \\
& =\frac{6}{3} \tan ^{-1}(3 x+2) \frac{-12}{-4} \sin ^{-1}(3-4 x)+c \\
& =2 \tan ^{-1}(3 x+2)+3 \sin ^{-1}(3-4 x)+c
\end{aligned}
$
Question 6 .
$
\frac{1}{3} \cos \left(\frac{x}{3}-4\right)+\frac{7}{7 x+9}+e^{\frac{x}{5}+3}
$
Solution:
$
\begin{aligned}
& =\int\left[\frac{1}{3} \cos \left(\frac{x}{3}-4\right)+\frac{7}{7 x+9}+e^{x / 5+3}\right] d x \\
& =\frac{1}{3} \int \cos \left(\frac{x}{3}-4\right) \mid x+7 \int \frac{1}{7 x+9} d x+\int e^{\frac{x}{5}+3} d x \\
& =\frac{1}{3} \frac{\sin \left(\frac{x}{3}-4\right)}{\frac{1}{3}}+\frac{7 \log |7 x+9|}{7}+\frac{e^{\frac{x}{5}+3}}{\frac{1}{5}}+c \\
& =\sin \left(\frac{x}{3}-4\right)+\log |7 x+9|+5 e^{\frac{x}{5}+3}+c
\end{aligned}
$