Exercise 11.2-Additional Questions - Chapter 11 - Integral Calculus - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
$
(3 x+4)^6
$
Solution:
$
\int(3 x+4)^6 d x=\frac{1}{3} \cdot \frac{(3 x+4)^{6+1}}{6+1} c=\frac{1}{3} \cdot \frac{(3 x+4)^7}{7}+c=\frac{(3 x+4)^7}{21}+c
$
Question 2.
$
\frac{1}{(x+5)^4}
$
Solution:
$
\int \frac{1}{(x+5)^4} d x=-\left(\frac{1}{1(4-1)}\right) \cdot \frac{1}{(x+5)^{4-1}}+c=\frac{-1}{3(x+5)^3}+c
$
Question 3.
$
\frac{1}{p+q x}
$
Solution:
$
\int \frac{1}{p+q x} d x=\frac{1}{q} \log (p+q x)+c
$
Question 4.
$
\cos (4 x+5)
$
Solution:
$
\int \cos (4 x+5) d x=\frac{\sin (4 x+5)}{4}=\frac{1}{4} \sin (4 x+5)+c
$

Question 5.
$\operatorname{cosec}^2(7-11 x)$

Solution:
$\int \operatorname{cosec}^2(7-11 x) d x=\frac{-\cot (7-11 x)}{-11}=\frac{1}{11} \cot (7-11 x)+c$
Question 6.
$\sec (3+x) \tan (3+x)$
Solution:
$
\int \sec (3+x) \tan (3+x) d x=\sec (3+x)+c
$
Question 7.
$\operatorname{cosec}(3-2 x) \cot (3-2 x)$
Solution:
$
\int \operatorname{cosec}(3-2 x) \cot (3-2 x) d x=\frac{-\operatorname{cosec}(3-2 x)}{-2}=\frac{1}{2} \operatorname{cosec}(3-2 x)+c
$
Question 8.
$
\mathrm{e}^{3 \mathrm{x}+2}
$
Solution:
$
\int e^{3 x+2} d x=\frac{e^{3 x+2}}{3}+c
$
Question 9.
$\frac{1}{\sin ^2(l-m x)}$
Solution:
$
\int \frac{1}{\sin ^2(l-m x)} d x=\int \operatorname{cosec}^2(l-m x) d x=\frac{-\cot (l-m x)}{-m}=\frac{1}{m} \cot (l-m x)+c
$
Question 10.
$
(\mathrm{l}+\mathrm{m})^{1 / 2}
$
Solution:

$\int(l x+m)^{\frac{1}{2}} d x=\frac{1}{l} \frac{(l x+m)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c=\frac{1}{\frac{3}{2} l}(l x+m)^{\frac{3}{2}}+c=\frac{2}{3 l}(l x+m)^{\frac{3}{2}}+c$