OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(d)

Students can track their progress and improvement through regular use of ISC Class 12 Maths Solutions OP Malhotra Chapter 15 Indefinite Integral-3 Ex 15(d).

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(d)

Question 1.
\(\int \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{1-x^2}}\)
Put x= sin θ ⇒ θ = sin^-1 x
⇒ dx = cos θ dθ
= \(\int \frac{\cos \theta d \theta}{\cos \theta}\) = θ + C = sin^-1 x + C

Question 2.
\(\int \frac{d x}{\sqrt{16-25 x^2}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{16-25 x^2}}\) = \(\frac{1}{5}\)\(\int \frac{d x}{\sqrt{\frac{16}{25}-x^2}}\)
= \(\frac{1}{5}\)\(\int \frac{d x}{\sqrt{\left(\frac{4}{5}\right)^2-x^2}}\)
= \(\frac{1}{5}\)sin^-1\(\left(\frac{x}{4 / 5}\right)\) + C
= \(\frac{1}{5}\)sin^-1\(\left(\frac{5 x}{4}\right)\) + C

Question 3.
\(\int \frac{d x}{\sqrt{4 x^2+9}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{4 x^2+9}}\) = \(\frac{1}{2}\)\(\int \frac{d x}{\sqrt{x^2+\left(\frac{3}{2}\right)^2}}\)
= \(\frac{1}{2}\)log\(\left|x+\sqrt{x^2+\left(\frac{3}{2}\right)^2}\right|+C\)
[∵ \(\int \frac{d x}{\sqrt{x^2+a^2}}\) = log|x + \(\sqrt{x^2+a^2}\)| + C]

Question 4.
\(\int \frac{d x}{\sqrt{9 x^2-25}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{9 x^2-25}}\)
= \(\frac{1}{3}\)\(\int \frac{2 x}{\sqrt{x^2-\left(\frac{5}{3}\right)^2}}\)
= \(\frac{1}{3}\)log|x + \(\sqrt{x^2-\left(\frac{5}{3}\right)^2}\)| + C
[∵ \(\int \frac{d x}{\sqrt{x^2-a^2}}\) = log | x + \(\sqrt{x^2-a^2}\) | + C ]

Question 5.
\(\int \frac{d x}{\sqrt{a^2-b^2 x^2}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{a^2-b^2 x^2}}\) = \(\frac{1}{b}\)\(\int \frac{d x}{\sqrt{\left(\frac{a}{b}\right)^2-x^2}}\)
= \(\frac{1}{b}\)sin^-1\(\left(\frac{x}{a / b}\right)\) + C
[∵ \(\int \frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{b}\)sin^-1\(\left(\frac{b x}{a}\right)\) + C

Question 6.
\(\int \frac{d x}{\sqrt{(2-x)^2+1}}\)
Solution:
Let I = \(\int \frac{d x}{\sqrt{(2-x)^2+1}}\);
Put 2 – x = t ⇒ -dx = dt
= \(\int \frac{-d t}{\sqrt{t^2+1^2}}\) = -log|t + \(\sqrt{t^2+1}\)| + C
= – log | 2 – x + \(\sqrt{(2-x)^2+1}\) | + C

Question 7.
\(\int \frac{x+2}{\sqrt{x^2+9}} d x\)
Solution:
Let I = \(\int \frac{x+2}{\sqrt{x^2+9}} d x\)
= \(\int \frac{x d x}{\sqrt{x^2+9}}\) + 2\(\int \frac{d x}{\sqrt{x^2+9}}\)
= \(\frac{1}{2}\)\(\int\left(x^2+9\right)\)^-1/2 2xdx + 2\(\int \frac{d x}{\sqrt{x^2+3^2}}\)

Question 8.
\(\int \frac{e^x d x}{\sqrt{4-e^{2 x}}}\)
Solution:
Let I = \(\int \frac{e^x d x}{\sqrt{4-e^{2 x}}}\);
Put e^x = t ⇒ e^x dx = dt
= \(\int \frac{d t}{\sqrt{4-t^2}}\) = \(\int \frac{d t}{\sqrt{2^2-t^2}}\)
= sin^-1\(\left(\frac{t}{2}\right)\) + C
[∵ \(\int \frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1\(\frac{x}{a}\) + C]
= sin^-1 \(\left(\frac{e^x}{2}\right)\) + C

Question 9.
\(\int \frac{x^2}{\sqrt{x^6-a^6}} d x\)
Solution:
Let I = \(\int \frac{x^2}{\sqrt{x^6-a^6}} d x\)
Put x³ = t ⇒ 3x²dx = dt
= \(\int \frac{d t}{3 \sqrt{t^2-\left(a^3\right)^2}}\)
= \(\frac{1}{3}\)log |t + \(\sqrt{t^2-a^6}\)| + C
= \(\frac{1}{3}\)log |x³ + \(\sqrt{x^6-a^6}\) | + C

Question 10.
\(\int \frac{\sec ^2 x}{\tan ^2 x+4} d x\)
Solution:
Let I = \(\int \frac{\sec ^2 x}{\tan ^2 x+4} d x\)
Put tan x = t ⇒ sec²x dx = dt
= \(\int \frac{d t}{t^2+4}\) = \(\int \frac{d t}{t^2+2^2}\) = \(\frac{1}{2}\)tan^-1\(\left(\frac{t}{2}\right)\) + C
[∵\(\int \frac{d x}{x^2+a^2}\) = \(\frac{1}{a}\)tan^-1\(\frac{x}{a}\) + C]
= \(\frac{1}{2}\)tan^-1\(\left(\frac{\tan x}{2}\right)\) + C
Or
Let I = \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x\)
Put tan x = t ⇒ sec²x dx = dt
= \(\int \frac{d t}{\sqrt{t^2+2^2}}=\log \left|t+\sqrt{t^2+4}\right|+C\)
= \(\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+C\)

Question 11.
\(\int \frac{a^x}{\sqrt{1-a^{2 x}}}\)
Solution:
Let I = \(\int \frac{a^x d x}{\sqrt{1-a^{2 x}}}\)
Put a^x = t ⇒ a^x log a dx = dt
= \(\int \frac{d t}{\log a \sqrt{1-t^2}}=\frac{1}{\log a} \sin ^{-1} t+C\)
= \(\frac{1}{\log a}\) sin^-1(a^x) + C

Question 12.
\(\int \sqrt{\frac{1-x}{1+x}} d x\)
Solution:
Let I = \(\int \sqrt{\frac{1-x}{1+x}} d x\)
= \(\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x\)
= \(\int \frac{1-x}{\sqrt{1-x^2}} d x\) = \(\int \frac{1}{\sqrt{1-x^2}} d x-\int \frac{x d x}{\sqrt{1-x^2}}\)
= sin^-1x + \(\frac{1}{2}\)\(\int\left(1-x^2\right)^{-\frac{1}{2}}(-2 x) d x\)
= sin^-1x + \(\frac{1}{2} \frac{\left(1-x^2\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\)
= sin^-1x + \(\sqrt{1-x^2}\) + C

Question 13.
\(\int \frac{(x-1)}{\sqrt{x^2-1}}\)
Solution:
Let I = \(\int \frac{(x-1)}{\sqrt{x^2-1}}\)
= \(\int \frac{x}{\sqrt{x^2-1}} d x\) – \(\int \frac{d x}{\sqrt{x^2-1}}\)
Put x² – 1 = t ⇒ 2x dx = dt
= \(\int \frac{d t}{2 \sqrt{t}}\) – log|x + \(\sqrt{x^2-1}\)| + C
= \(\frac{1}{2} \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) – log|x + \(\sqrt{x^2-1}\)| + C
= \(\sqrt{t}\) – log|x + \(\sqrt{x^2-1}\)| + C
= \(\sqrt{x^2-1}\) – log|x + \(\sqrt{x^2-1}\)| + C