ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.7

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.7

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 16) integrals:

Question 1.
(i) ∫ \(\frac{1}{x^2-16}\) dx (NCERT)
(ii) ∫ \(\frac{1}{9-x^2}\) dx
(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
(iv) ∫ \(\frac{d x}{x^2+16}\)
Solution:
(i) ∫ \(\frac{1}{x^2-16}\) dx
= (i) ∫ \(\frac{d x}{x^2-16}\)
= \(\frac{1}{2 \times 4} \log \left|\frac{x-4}{x+4}\right|\) + C
= \(\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(ii) ∫ \(\frac{1}{9-x^2}\) dx
= ∫ \(\frac{d x}{3^2-x^2}\)
= \(\frac{1}{2 \times 3} \log \left|\frac{3+x}{3-x}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3+x}{3-x}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{1^2-x^2}}\)
= sin^-1 \(\frac{x}{1}\) + C
= sin^-1 x + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]

(iv) ∫ \(\frac{d x}{x^2+16}\)
= ∫ \(\frac{d x}{x^2+4^2}\)
=\(\frac{1}{4}\) tan^-1 \(\frac{x}{4}\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

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

(ii) ∫ \(\frac{d x}{\sqrt{25+9 x^2}}\)
= \(\frac{1}{9} \int \frac{d x}{x^2+\left(\frac{5}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{\frac{5}{3}} \tan ^{-1}\left(\frac{x}{\frac{5}{3}}\right)\) + C
= \(\frac{1}{15} \tan ^{-1}\left(\frac{3 x}{5}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 2 (old).
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\) (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\)
= \(\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{3}\right)^2-x^2}}\)
= \(\frac{1}{3} \sin ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)\) + C

Question 3.
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
(ii) ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
Solution:
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
= \(\frac{1}{3} \int \frac{d x}{x \sqrt{x^2-\left(\frac{4}{3}\right)^2}}\)
= \(\frac{1}{3} \times \frac{1}{\frac{4}{3}} \sec ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
= \(\frac{1}{4} \sec ^{-1}\left(\frac{3 x}{4}\right)\) + C
[∵ ∫ \(\frac{d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+(\sqrt{5})^2}}\)
= log |x + \(\sqrt{x^2+5}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 3 (old).
(ii) ∫ \(\frac{d x}{\sqrt{x^2+4}}\)
Solution:
∫ \(\frac{d x}{\sqrt{x^2+4}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+2^2}}\)
= log |x + \(\sqrt{x^2+4}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 4.
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
(ii) ∫ \(\frac{d x}{32-2 x^2}\)
Solution:
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
= \(\frac{1}{9} \int \frac{d x}{x^2-\left(\frac{1}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{2 \times \frac{1}{3}} \log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3 x-1}{3 x+1}\right|\) + C

(ii) ∫ \(\frac{d x}{32-2 x^2}\)
= \(\frac{1}{2} \int \frac{d x}{16-x^2}\)
= \(\frac{1}{2} \int \frac{d x}{4^2-x^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 4} \log \left|\frac{4+x}{4-x}\right|\) + C
= \(\frac{1}{16} \log \left|\frac{4+x}{4-x}\right|\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\) (NCERT)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)

(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\)
= \(\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2+x^2}}\)
= \(\frac{1}{2} \log \left|x+\sqrt{x^2+\frac{1}{4}}\right|\) + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 6.
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy (ISC 2015)
(ii) ∫ \(\frac{x^4+1}{x^2+1}\) dx
Solution:
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy
= 2 ∫ \(\frac{y^2+4-4}{y^2+4}\) dy
= 2 ∫ [1 – \(\frac{4}{y^2+4}\)] dy
= 2 ∫ dy – 8 ∫ \(\frac{d y}{y^2+2^2}\)
= 2y – \(\frac{8}{2} \tan ^{-1}\left(\frac{y}{2}\right)\) + C
= 2y – 4 tan^-1 (\(\frac{y}{2}\)) + C

(ii) Let I = ∫ \(\frac{x^4+1}{x^2+1}\) dx
= ∫ [x² – 1 + \(\frac{2}{x^2+1}\)] dx
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 \(\frac{x}{1}\) + c
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 x + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + c]

Question 7.
(i) ∫ \(\frac{x}{x^4-1}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x^2}{1+x^6}\) dx (NCERT)
Solution:
(i) I = ∫ \(\frac{x}{x^4-1}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\int \frac{d t}{2\left(t^2-1\right)}\)
= \(\frac{1}{2} \int \frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \mid \frac{x-a}{x+a}\) + C]

(ii) Let I = \(\frac{3 x^2}{1+x^6}\) dx
∴ I = ∫ \(\frac{d t}{t^2+1}\)
= tan^-1 t + C
= tan^-1 (x³) + C

Question 8.
(i) ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx (NCERT)
(ii) ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx
put x³ = t
⇒ 3x² dx = dt
∴ I = ∫ \(\frac{d t}{3 \sqrt{t^2+\left(a^3\right)^2}}\)
= \(\frac{1}{3}\) log |t + \(\sqrt{t^2+a^6}\)| + C
= \(\frac{1}{3}\) log |x³ + \(\sqrt{x^6+a^6}\)| + C

(ii) I = ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx
put x⁴ = t
⇒ 4x³ dx = dt
∴ I = \(\int \frac{d t}{4 \sqrt{1-t^2}}\)
= \(\frac{1}{4} \sin ^{-1} \frac{t}{1}\) + C
= \(\frac{1}{4}\) sin^-1 (x⁴) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]

Question 9.
(i) ∫ \(\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x}{1+2 x^4}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\sqrt{x} d x}{\sqrt{a^3-x^3}}\)

(ii) Let I = ∫ \(\frac{3 x}{1+2 x^4}\) dx
= ∫ \(\frac{3 x}{1+2\left(x^2\right)^2}\) dx
put x² = t
∴ 2x dx = dt
∴ I = 3 ∫ \(\frac{d t}{2\left[1+2 t^2\right]}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\frac{1}{2}}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2}\)
= \(\frac{3}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\) + C
= \(\frac{3 \sqrt{2}}{4}\) tan^-1 (√2 x²) + C

Question 10.
(i) ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx (NCERT)
(ii) ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{d t}{\sqrt{4-t^2}}\)
= ∫ \(\frac{d t}{\sqrt{2^2-t^2}}\)
= sin^-1 (\(\frac{t}{2}\)) + c
= sin^-1 (\(\left(\frac{\sin x}{2}\right)\)) + c

(ii) Let I = ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
put tan x = t
⇒ ∫ sec² x dx = dt
= ∫ \(\frac{d t}{\sqrt{4+t^2}}\)
= ∫ \(\frac{d t}{\sqrt{t^2+2^2}}\)
= log |t + \(\sqrt{t^2+4}\)| + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + c
= log |tan x + \(\sqrt{\tan ^2 x+4}\)| + c

Question 11.
(i) ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
(ii) ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
put tan³ x = t
⇒ 3 tan² x sec² x dx = dt
∴ I = ∫ \(\frac{d t}{3\left(1-t^2\right)}\)
= \(\frac{1}{3} \times \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|\) + C
[∵ \(\frac{d x}{a^2+x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]
= \(\frac{1}{6} \log \left|\frac{1+\tan ^3 x}{1-\tan ^3 x}\right|\) + C

Question 11 (old).
(ii) ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
Solution:
Let I = ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
put cos x = t
⇒ – cosec² x dx = dt
∴ I = ∫ \(\frac{-d t}{1-t^2}\)
= – \(\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\)
= – \(\frac{1}{2} \log \left|\frac{1+\cot x}{1-\cot x}\right|\) + C
[∵ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 12.
(i) ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
(ii) ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
put sin x – cos x = t ……….(1)
⇒ (cos x + sin x) dx = dt
On squaring (1) ; we have
sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
∴ I = ∫ \(\frac{d t}{\sqrt{1-t^2}}\)
= sin^-1 (t) + C
= sin^-1 (sin x – cos x) + C

(ii) Let I = ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Divide Numerator and denominator by cos² x ; we get
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ \(\frac{d t}{2 t^2+5}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\left(\sqrt{\frac{5}{2}}\right)^2}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{\frac{5}{2}}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{2}}}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]
= \(\frac{1}{\sqrt{10}} \tan ^{-1}\left(\sqrt{\frac{2}{5}} \tan x\right)\) + C

Question 12 (old).
(ii) ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
Solution:
Let I = ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx

Question 13.
(i) ∫ \(\frac{\cos x}{\cos 3 x}\) dx
(ii) ∫ \(\frac{d x}{2-3 \cos 2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\cos 3 x}\) dx

(ii) Let I = ∫ \(\frac{d x}{2-3 \cos 2 x}\)
= ∫ \(\frac{d x}{2-3\left(1-2 \sin ^2 x\right)}\)
= ∫ \(\frac{d x}{-1+6 \sin ^2 x}\)
Divide numerator and denominator by cos² x ; we have
= ∫ \(\frac{\sec ^2 x d x}{-\sec ^2 x+6 \tan ^2 x}\)
=∫ \(\frac{\sec ^2 x d x}{-1-\tan ^2 x+6 \tan ^2 x}\)

Question 13 (old).
(ii) ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Solution:
Let I = ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Divide numerator and denominator by cos² x ; we have

Question 14.
(i) ∫ \(\frac{d x}{e^x-e^{-x}}\)
(ii) ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\frac{d x}{e^x-e^{-x}}\)
= ∫ \(\frac{d x}{e^x-\frac{1}{e^x}}\)
= ∫ \(\frac{e^x d x}{e^{2 x}-1}\)
put e^x = t
⇒ e^x dx = dt
= ∫ \(\frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right|\) + C

(ii) Let I = ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx
put x = cos θ
⇒ dx = – sin θ dθ

= – [θ + sin θ] + C
= – cos^-1 x – \(\sqrt{1-x^{2}}\) + C
[∵ cos θ = x
⇒ θ = cos^-1 x
and sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-x^2}\)]

Question 15.
(i) ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx (NCERT)
(ii) ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}-\int \frac{d x}{\sqrt{x^2-1}}\)
= I₁ – \(\frac{d x}{\sqrt{x^2-1}}\) …..(1)
where I₁ = ∫ \(\frac{x d x}{\sqrt{x^2-1}}\) ……….(1)
put x² = t
⇒ 2x dx = dt
∴ I₁ = ∫ \(\frac{d t}{2 \sqrt{t-1}}\)
= \(\frac{1}{2}\) (t – 1)^-1/2 dt
= \(\frac{1}{2} \frac{(t-1)^{1 / 2}}{1 / 2}\)
= \(\sqrt{x^{2}-1}\)
I = \(\sqrt{x^2-1}-\log \left|x+\sqrt{x^2-1}\right|\) + C

(ii) Let I = ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}+2 \int \frac{d x}{\sqrt{x^2-1}}\)
= \(\frac{1}{2}\) ∫ (x^{2} – 1)^{\(-\frac{1}{2}\)} (2x dx) + 2 ∫ \(\frac{d x}{\sqrt{x^2-1^2}}\)
= \(\frac{1}{2} \frac{\left(x^2-1\right)^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{(f(x))^{n+1}}{n+1}\) + C]
= \(\sqrt{x^2-1}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C

Question 16.
(i) ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{\sqrt{x^2-a^2}}{x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx
∴ I = ∫ \(\sqrt{\frac{(a+x)}{a-x} \times \frac{a+x}{a+x}}\) dx
= ∫ \(\frac{a+x}{\sqrt{a^2-x^2}}\) dx
= ∫ \(a \int \frac{d x}{\sqrt{a^2-x^2}}+\int \frac{x d x}{\sqrt{a^2-x^2}}\)
∴ I = a sin^-1 \(\frac{x}{a}\) + I₁
where I₁ = ∫ \(\frac{d t}{2 \sqrt{a^2-t}}\)
= \(\frac{1}{2}\) ∫ (a² – t)^-1/2 dt
= – \(\frac{1}{2} \frac{\left(a^2-t\right)^{1 / 2}}{1 / 2}\)
= – \(\sqrt{a^2-x^2}\)
∴ From (1) ; we have
∴ I = a sin^-1 \(\frac{x}{a}\) – \(\sqrt{a^2-x^2}\) + C

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

Question 17.
If ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin^-1 (2^x) + C, then what is the value of k?
Solution:
Let I = ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx
= ∫ \(\frac{2^x d x}{\sqrt{1-\left(2^x\right)^2}}\)
put 2^x = t
⇒ 2^x log 2 dx = dt
⇒ I = ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\)
= \(\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^2}}\)
I = \(\frac{1}{\log 2}\) sin^-1 t + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C
∴ I = \(\frac{1}{\log 2}\) sin^-1 (2^x) + C ………….(1)
Also given
∫ \(\frac{2^x d x}{1-4 x}\) = k sin^-1 (2^x) + C ………..(2)
From (1) and (2) ; we have
k = \(\frac{1}{\log 2}\)