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

Practicing ISC Class 12 Maths Solutions OP Malhotra Chapter 15 Indefinite Integral-3 Ex 15(f) is the ultimate need for students who intend to score good marks in examinations.

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

Question 1.
\(\int \sqrt{1-x^2} d x\)
Solution:
\(\int \sqrt{1-x^2} d x ;\)
Put x = sin θ ⇒ dx = cos θ dθ
= \(\int \sqrt{1-\sin ^2 \theta} \cos d \theta\)
= \(\int \cos ^2 \theta d \theta=\int\left[\frac{1+\cos 2 \theta}{2}\right] d \theta\)
= \(\frac { 1 }{ 2 }\)\(\left[\theta+\frac{\sin 2 \theta}{2}\right]\)
= \(\frac { 1 }{ 2 }\)sin^-1x + \(\frac { 1 }{ 2 }\)\(\sqrt{1-x^2}\) + C
Aliter : \(\int \sqrt{1^2-x^2} d x\)
= \(\frac{x \sqrt{1-x^2}}{2}\) + \(\frac { 1 }{ 2 }\)sin^-1\(\frac { x }{ 1 }\) + C
[∵ \(\int \sqrt{a^2-x^2} d x\) = \(\frac{x \sqrt{a^2-x^2}}{2}\) + \(\frac{a^2}{2}\)sin^-1\(\left(\frac{x}{a}\right)\) + C]

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

Question 3.
\(\int \sqrt{3-4 x^2} d x\)
Solution:
\(\int \sqrt{3-4 x^2} d x\) = \(2 \int \sqrt{\frac{3}{4}-x^2} d x\)

Question 4.
\(\int \sqrt{1+x^2} d x\)
Solution:
\(\int \sqrt{1+x^2} d x\) = \(\frac{x \sqrt{1+x^2}}{2}\) + \(\frac { 1 }{ 2 }\)log|x + \(\sqrt{1+x^2}\)| + C

Question 5.
\(\int \sqrt{16+x^2} d x\)
Solution:
\(\int \sqrt{16+x^2} d x\) = \(\int \sqrt{4^2+x^2} d x\)
= \(\frac{x \sqrt{16+x^2}}{2}\) + \(\frac { 16 }{ 2 }\)log|x + \(\sqrt{16+x^2}\)| + C = \(\frac{x \sqrt{16+x^2}}{2}\) + 8log|x + \(\sqrt{16+x^2}\)| + C

Question 6.
\(\int \sqrt{x^2-36} d x\)
Solution:

Question 7.
\(\int \sqrt{3 x^2+5} d x\)
Solution:
\(\int \sqrt{3 x^2+5} d x\) = \(\sqrt{3} \int \sqrt{x^2+\frac{5}{3}} d x\) = \(\sqrt{3} \int \sqrt{x^2+\left(\sqrt{\frac{5}{3}}\right)^2} d x\)

Question 8.
\(\int x \sqrt{x^4+1} d x\)
Solution:
I = \(\int x \sqrt{x^4+1} d x\);
Put x² = t ⇒ 2xdx = dt

Question 9.
\(\int \frac{\sqrt{1+(\log x)^2}}{x} d x\)
Solution:
Let I = \(\int \frac{\sqrt{1+(\log x)^2}}{x} d x\);
Put log x = t ⇒ 1/x dx = dt

Question 10.
\(\int \sqrt{x^2+4 x+6} d x\)
Solution:
Let I = \(\int \sqrt{x^2+4 x+6} d x\)
= \(\int \sqrt{x^2+4 x+4+2} d x\) = \(\int \sqrt{(x+2)^2+(\sqrt{2})^2} d x\)
Put x + 2 = t ⇒ dx = dt
= \(\int \sqrt{t^2+(\sqrt{2})^2} d t\) = \(\frac{t \sqrt{t^2+2}}{2}\) + \(\frac { 2 }{ 2 }\)log|t + \(\)| + C
= \(\frac{(x+2) \sqrt{x^2+4 x+6}}{2}\) + log|x + 2 + \(\sqrt{x^2+4 x+6}\)| + C

Question 11.
\(\int \sqrt{x^2+3 x} d x\)
Solution:
Let I = \(\int \sqrt{x^2+3 x} d x\)

Question 12.
\(\int \sqrt{1+2 x-3 x^2} d x\)
Solution:

Question 13.
\(\int \sqrt{1-4 x-x^2} d x\)
Solution:
Let I = \(\int \sqrt{1-4 x-x^2} d x\)
= \(\int \sqrt{-\left(x^2+4 x+4-5\right)} d x\) = \(\int \sqrt{5-(x+2)^2} d x\) = \(\int \sqrt{(\sqrt{5})^2-(x+2)^2} d x ;\)
Put x + 2 = t ⇒ dx = dt
= \(\int \sqrt{(\sqrt{5})-t^2} d t\) = \(\frac{t \sqrt{5-t^2}}{2}\) + \(\frac { 5 }{ 2 }\)sin^-1\(\frac{t}{\sqrt{5}}\) + C
= \(\frac{(x+2) \sqrt{1-4 x-x^2}}{2}\) + \(\frac { 5 }{ 2 }\)sin^-1\(\left(\frac{x+2}{\sqrt{5}}\right)\) + C

Question 14.
\(\int \sqrt{x(1-x)} d x\)
Solution:
Let I = \(\int \sqrt{x(1-x)} d x\) = \(\int \sqrt{x-x^2} d x\)

Question 15.
\(\int \sqrt{x^2+4 x+1} d x\)
Solution:
Let I = \(\int \sqrt{x^2+4 x+1} d x\)
= \(\int \sqrt{x^2+4 x+4-3} d x\) = \(\int \sqrt{(x+2)^2-(\sqrt{3})^2} d x\)
= \(\frac{(x+2) \sqrt{(x+2)^2-3}}{2}\) – \(\frac{3}{2}\)log|x + 2 + \(\sqrt{(x+2)^2-3}\)| + C
[∵ \(\int \sqrt{x^2-a^2} d x\) = \(\frac{x \sqrt{x^2-a^2}}{2}\) – \(\frac{a^2}{2}\)log|x + \(\sqrt{x^2-a^2}\)| + C]
= \(\frac{(x+2) \sqrt{x^2+4 x+1}}{2}\) – \(\)log|x + 2 +\(\)| + C

Question 16.
\(\int(2 x+3) \sqrt{x^2+4 x+3} d x\)
Solution:
Let I = \(\int(2 x+3) \sqrt{x^2+4 x+3} d x\)
= \(\int(2 x+4-1) \sqrt{x^2+4 x+3} d x\)
= \(\int(2 x+4) \sqrt{x^2+4 x+3} d x\) – \(\int \sqrt{x^2+4 x+3} d x\) = \(\int \sqrt{t} d t\) – \(\int \sqrt{x^2+4 x+4-1} d x\)
[Put x² + 4x + 3 = t ⇒ (2x + 4) dx = dt]
= \(\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) – \(\int \sqrt{(x+2)^2-1^2} d x\)
= \(\frac{2}{3} t^{3 / 2}\) – [\(\frac{(x+2) \sqrt{(x+2)^2-1}}{2}\) – \(\frac { 1 }{ 2 }\)log | x + 2 + \(\sqrt{(x+2)^2-1}\)|] + C
= \(\frac{2}{3}\left(x^2+4 x+3\right)^{\frac{3}{2}}\) – \(\left(\frac{x+2}{2}\right) \sqrt{x^2+4 x+3}\) + \(\frac { 1 }{ 2 }\)log|x + 2 + \(\sqrt{(x+2)^2-1}\)| + C

Question 17.
\(\int(2 x-5) \sqrt{x^2-4 x+3} d x\)
Solution:
Let I = \(\int(2 x-5) \sqrt{x^2-4 x+3} d x\)
= \(\int(2 x-4-1) \sqrt{x^2-4 x+3} d x\) = \(\int \sqrt{x^2-4 x+3}(2 x-4) d x\) – \(\int \sqrt{x^2-4 x+3} d x\)
Put x² + 4x + 3 = t ⇒ (2x – 4) dx = dt

Question 18.
\(\int x \sqrt{x+x^2} d x\)
Solution:
Let I = \(\int x \sqrt{x+x^2} d x\)
= \(\frac { 1 }{ 2 }\)\(\int(2 x+1-1)\)\(\sqrt{x+x^2} d x\) = \(\frac { 1 }{ 2 }\)\(\int\left(x+x^2\right)^{\frac{1}{2}}(2 x+1) d x\) – \(\frac{1}{2} \int \sqrt{x^2+x} d x\)
Put x + x² = t ⇒ (1 + 2x) dx = dt

Question 19.
\(\int(x-5) \sqrt{x^2+x} d x\)
Solution:
Let I = \(\int(x-5) \sqrt{x^2+x} d x\)
= \(\frac { 1 }{ 2 }\)(2x + 10)\(\sqrt{x^2+x} d x\) = \(\frac { 1 }{ 2 }\)\(\int(2 x+1-11) \sqrt{x^2+x} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \sqrt{x^2+x}(2 x+1) d x\) – \(\frac { 11 }{ 2 }\)\(\int \sqrt{x^2+x} d x\)
= \(\frac { 1 }{ 2 }\)\(\int \sqrt{t} d t\) – \(\frac { 11 }{ 2 }\)\(\int \sqrt{x^2+x+\frac{1}{4}-\frac{1}{4}} d x\)
Put x² + x = t ⇒ (2x + 1) dx = dt

Question 20.
\(\int(x+3) \sqrt{3-4 x-x^2} d x\)
Solution:
Let I = \(\int(x+3) \sqrt{3-4 x-x^2} d x\)
= – \(\frac { 1 }{ 2 }\)\(\int(-2 x-6) \sqrt{3-4 x-x^2} d x\) = –\(\frac { 1 }{ 2 }\)\(\int(-2 x-4-2) \sqrt{3-4 x-x^2} d x\)
= – \(\frac { 1 }{ 2 }\)\(\int \sqrt{3-4 x-x^2}(-2 x-4) d x\) + \(\int \sqrt{3-4 x-x^2} d x\)
= – \(\frac { 1 }{ 2 }\)\(\int \sqrt{t} d t\) + \(\int \sqrt{-\left(x^2+4 x-3\right)} d x\)
[Put 3 – 4x – x² = t ⇒ (-4 – 2x)dx = dt]