The value of the integral $\int \frac{\sin \theta \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ (where c is a constant of integration)

(B) $\frac{1}{18}\left[11-18 \cos ^{2} \theta+9 \cos ^{4} \theta-2 \cos ^{6} \theta\right]^{\frac{3}{2}}+c$

(A) $\frac{1}{18}\left[9-2 \sin ^{6} \theta-3 \sin ^{4} \theta-6 \sin ^{2} \theta\right]^{\frac{3}{2}}+c$

(C) $\frac{1}{18}\left[11-18 \sin ^{2} \theta+9 \sin ^{4} \theta-2 \sin ^{6} \theta\right]^{\frac{3}{2}}+c$

(D) $\frac{1}{18}\left[9-2 \cos ^{6} \theta-3 \cos ^{4} \theta-6 \cos ^{2} \theta\right]^{\frac{3}{2}}+c$

Indefinite Integrals — JEE Main · 2021 · 25 Feb 2021 · Shift 1 Question with Solution

The value of the integral

\int \frac{s i n θ s i n 2 θ ( s i n ^{6} θ + s i n ^{4} θ + s i n ^{2} θ ) 2 s i n ^{4} θ + 3 s i n ^{2} θ + 6}{1 - c o s 2 θ} d θ

(where c is a constant of integration)

(A) $\frac{1}{18} [9 - 2 sin^{6} θ - 3 sin^{4} θ - 6 sin^{2} θ]^{\frac{3}{2}} + c$
(B) $\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{\frac{3}{2}} + c$ ✓ Correct
(C) $\frac{1}{18} [11 - 18 sin^{2} θ + 9 sin^{4} θ - 2 sin^{6} θ]^{\frac{3}{2}} + c$
(D) $\frac{1}{18} [9 - 2 cos^{6} θ - 3 cos^{4} θ - 6 cos^{2} θ]^{\frac{3}{2}} + c$

Answer: (B)

Since

1 - cos 2 θ = 2 sin^{2} θ

and

sin 2 θ = 2 sin θ cos θ

, the integrand reduces to

cos θ (sin^{6} θ + sin^{4} θ + sin^{2} θ) 2 sin^{4} θ + 3 sin^{2} θ + 6 .

Taking one factor

sin^{2} θ

inside the root, with

sin^{6} θ + sin^{4} θ + sin^{2} θ = sin^{2} θ (sin^{4} θ + sin^{2} θ + 1)

and

sin^{2} θ (2 sin^{4} θ + 3 sin^{2} θ + 6) = 2 sin^{6} θ + 3 sin^{4} θ + 6 sin^{2} θ

, the integrand equals

sin θ cos θ (sin^{4} θ + sin^{2} θ + 1) 2 sin^{6} θ + 3 sin^{4} θ + 6 sin^{2} θ .

Putting

t = 2 sin^{6} θ + 3 sin^{4} θ + 6 sin^{2} θ

gives

d t = 12 sin θ cos θ (sin^{4} θ + sin^{2} θ + 1) d θ

, so the integral is

\frac{1}{12} \int t d t = \frac{t ^{3/2}}{18} + c

. Replacing

sin^{2} θ

1 - cos^{2} θ

gives

t = 11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ

, so the integral equals

\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{3/2} + c

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option