OP Malhotra Class 12 Maths Solutions Chapter 13 Indefinite Integral-1 Ex 13(a)

Utilizing ISC Class 12 Maths Solutions OP Malhotra Chapter 13 Indefinite Integral-1 Ex 13(a) as a study aid can enhance exam preparation.

S Chand Class 12 ICSE Maths Solutions Chapter 13 Indefinite Integral-1 Ex 13(a)

Question 1.
(i) sin 2x
(ii) 2 sin 3x
(iii) \(\frac { 1 }{ 3 }\)cos4x
(iv) \(\frac{\cos 5 x}{2}\)
(v) 8 cos² 8
(vi) cosec² 2x
(vii) sec 5x tan 5x
(viii) -cosec 3x cot 3x
Solution:

Question 2.
(i) cos (5 – 3x)
(ii) 2sin \(\left(\frac{\pi}{2}-\frac{x}{2}\right)\)
(iii) sin\(\left(\frac{3}{4} x+5\right)\)
(iv) 4 sec²(2x – 4)
Solution:

Question 3.
(i) sin² x
(ii) cos² x
(iii) sin³ x
(iv) sin² mx
(v) sin² x cos² x
(vi) sin³ x cos³ x
(vii) \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\)
(viii) sin x sec² x
(ix) sin³ x cos³ x
(x) \(\frac{1}{\sin ^2 x \cos ^2 x}\)
(xi) \(\frac{\sec x}{\sec +\tan x}\)
(xii) 3 cosec² x + 2 sin3x
Solution:

Question 4.
(i) cos 4x cos 3x dx
(ii) sin 4xsin 8x
Solution:

Question 5.
(i) cos2xcos4xcos6x
(ii) sin x sin2xsin3x
(iii) \(\frac{\cos ^2 x-\sin ^2 x}{\sqrt{1+\cos 4 x}}\)
(iv) cos⁴ xsin⁴x
(v) \(\frac{7 \cos ^3 x+4 \sin ^3 x}{3 \sin ^2 x \cos ^2 x}\)
Solution:

Question 6.
(i) \(\frac{1}{1+\cos x}\)
(ii) \(\frac{1}{1-\cos 2 x}\)
(iii) \(\frac{1}{1-\sin x}\)
(iv) \(\frac{1-\cos 2 x}{1-\cos 2 x}\)
(v) \(\sqrt{1+\cos x}\)
(vi) \(\sqrt{1+\sin 2 x}\)
(vii) \(\cos x \sqrt{1+\cos 2 x}\)
(viii) \(\sin x \sqrt{1-\cos 2 x}\)
(ix) \(\frac{\cos x-\sin x}{\cos x+\sin x}(2+2 \sin 2 x)\)
(x) \(\frac{4-5 \sin x}{\cos ^2 x}+\frac{1}{\sin ^2 x \cos ^2 x}\)
(xi) \(\frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}\)
Solution:

Question 7.
\(\sqrt{\left(1+\sin \frac{x}{2}\right)}\)
Solution:

Question 8.
\(\frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x}\)
Solution:

Question 9.
sin⁶x
Solution:

Question 10.
tan^-1\(\left(\frac{\sin 2 x}{1+\cos 2 x}\right)\)
Solution:

Question 11.
cos^-1\(\left(\frac{1-\tan ^2 x}{1+\tan x}\right)\)
Solution:

Question 12.
cos^-1(sin x)
Solution:

Question 13.
If f ‘ (x) = 3x² – \(\frac{2}{x^3}\) and f (1) = 0, find f(1).
Solution:
Given f ‘ (x) = 3x² – \(\frac{2}{x^3}\);
on integrating both sides, we have
f(x) = \(\frac{3 x^3}{3}\) – 2\(\frac{x^{-3+1}}{-3+1}\) + C
⇒ f(x) = x³ + \(\frac{1}{x^2}\) + C …(1)
since f(1) = 0 i.e. when x = 1, f(x) = 0
∴ from (1) ; 0 = 1 + \(\frac{1}{1}\) + C
⇒ C = -2
∴ eqn (1) gives ; f(x) = x³ + \(\frac{1}{x^2}\) – 2

Question 14.
If f ‘ (x) = a sin x + b cos x and f ‘ (0) = 4, f(0) = 3, f\(\left(\frac{\pi}{2}\right)\) = 5, find f(x).
Solution:
Gives f ‘ (x) = a sin x + b cos x
Since f ‘ (0) = 4
∴ from (1) ; 4 = a × 0 + b × 1
⇒ b = 4
Also f(x) = \(\int f^{\prime}(x) d x+C\)
⇒ f(x) = \(\int(a \sin x+b \cos x) d x+\mathrm{C}\)
⇒ f(x) = – a cos x + b sin x + C …(2)
since f(0) = 3 i.e. when x = 0; f(x) = 3
∴ from (2); 3 = – a \times 1 + b × 0 + C
⇒ 3 = – a + C
Also f(π/2) = 5 i.e. when x = (π/2),(x) = 5
∴ from (2); 5 = – a × 0 + b × 1 + C
⇒ 5 = b + C ⇒ 5 = 4 + C
⇒ C = 1
∴ from (3); 3 = – a + 1 ⇒ a = – 2
∴ from (2); we have
f(x) = – 2 cos x + 4 sin x + 1