Exercise 11.2 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.2
Integrate the following functions w.r.to ' $\mathrm{x}$ '.
Question 1.
(i) $(x+5)^6$
(ii) $\frac{1}{(2-3 x)^4}$
(iii) $\sqrt{3 x+2}$
Solution:
(i) $\int(x+5)^6 d x=\frac{(x+5)^7}{7}+c$
(ii) $\int \frac{1}{(2-3 x)^4} d x=\frac{-1}{-3 \times 3(2-3 x)^3}=\frac{1}{9(2-3 x)^3}+c$
(iii) $\int \sqrt{3 x+2} d x=\int(3 x+2)^{1 / 2} d x$
$
\begin{aligned}
& =\frac{1}{3} \frac{(3 x+2)^{3 / 2}}{3 / 2} \\
& =\frac{2}{9}(3 x+2)^{3 / 2}+c
\end{aligned}
$
Question 2.
(i) $\sin 3 \mathrm{x}$
(ii) $\cos (5-11 x)$
(iii) $\operatorname{cosec}^2(5 \mathrm{x}-7)$
Solution:
(i) $\int \sin 3 x d x=\frac{-1}{3} \cos 3 x+c$
(ii) $\int \cos (5-11 x) d x=\frac{-1}{11} \sin (5-11 x)+c$
(iii) $\int \operatorname{cosec}^2(5 x-7) d x=\frac{-1}{5} \cot (5 x-7)+c$

Question 3.
(i) $\mathrm{e}^{3 \mathrm{x}-6}$
(ii) $\mathrm{e}^{8-7 \mathrm{x}}$
(iii) $\frac{1}{6-4 x}$
Solution:
(i) $\int e^{3 x-6} d x=\frac{1}{3} e^{3 x-6}+c$
(ii) $\int e^{8-7 x} d x=\frac{-1}{7} e^{8-7 x}+c$
(iii) $\int \frac{1}{6-4 x} d x=\frac{-1}{4} \log (6-4 x)+c$
Question 4.
(i) $\sec ^2 \frac{x}{5}$
(ii) $\operatorname{cosec}(5 x+3) \cot (5 x+3)$
(iii) $\sec (2-15 x) \tan (2-15 \mathrm{x})$
Solution:
(i) $\int \sec ^2 \frac{x}{5} d x=5 \tan \frac{x}{5}+c$
(ii) $\int \operatorname{cosec}(5 x+3) \cot (5 x+3) d x=\frac{-1}{5} \operatorname{cosec}(5 x+3)+c$
(iii) $\int \sec (2-15 x) \tan (2-15 x) d x$
$
\begin{aligned}
& =\frac{\sec (2-15 x)}{-15}+c \\
& =-\frac{1}{15} \sec (2-15 x)+c
\end{aligned}
$

Question 5.
(i) $\frac{1}{\sqrt{1-(4 x)^2}}$
(ii) $\frac{1}{\sqrt{1-81 x^2}}$
(iii) $\frac{1}{1+36 x^2}$
Solution:
(i) $\int \frac{1}{\sqrt{1-(4 x)^2}} d x=\frac{1}{4} \sin ^{-1}(4 x)+c$
(ii) $\int \frac{1}{\sqrt{1-81 x^2}} d x=\int \frac{1}{\sqrt{1-(9 x)^2}} d x=\frac{1}{9} \sin ^{-1}(9 x)+c$
(iii) $\int \frac{1}{1+36 x^2} d x=\int \frac{1}{1+(6 x)^2} d x=\frac{1}{6} \tan ^{-1}(6 x)+c$