ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.13

Utilizing ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals Ex 8.13 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.13

Very Short answer type questions (1 to 3) :

Evaluate the following (1 to 8) integrals :

Question 1.
(i) ∫ \(\sqrt{4-x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{x^2-9}\) dx
Solution:
(i) ∫ \(\sqrt{4-x^2}\) dx

(ii) ∫ \(\sqrt{x^2-9}\) dx

Question 2.
(i) ∫ \(\sqrt{1+x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{1-4 x^2}\) dx (NCERT)
Solution:
(i) ∫ \(\sqrt{1+x^2}\) dx

(ii) ∫ \(\sqrt{1-4 x^2}\) dx

Question 3.
(i) ∫ \(\sqrt{4 x^2-9}\) dx
(ii) ∫ \(\sqrt{4-9 x^2}\) dx
Solution:
(i) ∫ \(\sqrt{4 x^2-9}\) dx

(ii) ∫ \(\sqrt{4-9 x^2}\) dx

Question 4.
(i) ∫ \(\frac{x^2}{\sqrt{x^2+6}}\) dx
(ii) ∫ \(\frac{1}{x-\sqrt{x^2-1}}\) dx
Solution:
(i) ∫ \(\frac{x^2}{\sqrt{x^2+6}}\) dx

(ii) ∫ \(\frac{d x}{x-\sqrt{x^2-1}}\)

Question 5.
(i) ∫ \(\frac{x^2+3}{\sqrt{x^2-25}}\) dx
(ii) ∫ (x – 3) \(\sqrt{\frac{x+2}{x-2}}\) dx
Solution:
(i) ∫ \(\frac{x^2+3}{\sqrt{x^2-25}}\) dx

(ii) Let I = ∫ (x – 3) \(\sqrt{\frac{x+2}{x-2}}\) dx

Question 6.
(i) ∫ x \(\sqrt{x^4-1}\) dx
(ii) ∫ \(\frac{\sqrt{9-(\log x)^2}}{x}\) dx
Solution:
(i) Let I = ∫ x \(\sqrt{x^4-1}\) dx ;
put x² = t
⇒ x dx = \(\frac{d t}{2}\)

(ii) Let I = ∫ \(\frac{\sqrt{9-(\log x)^2}}{x}\) dx

put log x = t
⇒ \(\frac{1}{x}\) dx = dt

Question 7.
(i) ∫ sin x cos x \(\sqrt{\sin ^4 x+4}\) dx
(ii) ∫ \(\frac{\log x}{x} \sqrt{(\log x)^4-1}\) dx
Solution:
(i) Let I = ∫ sin x cos x \(\sqrt{\sin ^4 x+4}\) dx
put sin² x = t
⇒ 2 sin x cos x dx = dt

(ii) Let I = ∫ \(\frac{\log x}{x} \sqrt{(\log x)^4-1}\) dx
put (log x)² = t
⇒ 2 log x . \(\frac{1}{x}\) dx = dt
= ∫ \(\sqrt{t^2-1^2} \frac{d t}{2}\)
= \(\frac{1}{2}\left[\frac{t \sqrt{t^2-1}}{2}-\frac{1}{2} \log \left|t+\sqrt{t^2-1}\right|\right]\)
= \(\frac{1}{4}\left[(\log x)^2 \sqrt{(\log x)^4-1}-\log \left|(\log x)^2+\sqrt{(\log x)^4-1}\right|\right]\) + C

Question 8.
(i) ∫ x cos^-1 x dx (NCERT)
(ii) ∫ cos^-1 √x dx
Solution:
(i) ∫ x cos^-1 x dx

(ii) Let I = ∫ cos^-1 √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt