The value of the integral

$\int \frac{s i n θ s i n θ (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{1 - c o s 2 θ} d θ$ (where c is a constant of integration)

Option 1 -

$\frac{1}{18} {[9 - 2 s i n^{6} θ - 3 s i n^{4} θ - 6 s i n^{2} θ]}^{\frac{3}{2}} + c$

Option 2 -

$\frac{1}{18} {[11 - 18 c o s^{2} θ + 9 c o s^{4} θ - 2 c o s^{6} θ]}^{\frac{3}{2}} + c$

Option 3 -

$\frac{1}{18} {[11 - 18 s i n^{2} θ + 9 s i n^{4} θ - 2 s i n^{6} θ]}^{\frac{3}{2}} + c$

Option 4 -

$\frac{1}{18} {[9 - c o s^{6} θ - 3 c o s^{4} θ - 6 c o s^{2} θ]}^{\frac{3}{2}} + c$

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

Let sin = t

$\int \frac{s i n θ (2 s i n θ . c o s θ) (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{2 s i n^{2} θ}$ d

sin = t

cos . d = dt

$\int \frac{u^{1 / 2}}{12} d u = \frac{u^{3 / 2}}{18} + C = \frac{{(2 t^{6} + 3 t^{4} + 6 t^{2})}^{3 / 2}}{18} + C = \frac{{(2 s i n^{6} θ + 3 s i n^{4} θ + 6 s i n^{2} θ)}^{3 / 2}}{18} + C$

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The value of the integral

$\int \frac{s i n θ s i n θ (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{1 - c o s 2 θ} d θ$ (where c is a constant of integration)

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The value of the integral ∫ s i n θ s i n 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)

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The value of the integral

$\int \frac{s i n θ s i n θ (s i n^{6} θ + s i n^{4} θ + s i n^{2} θ) \sqrt[]{2 s i n^{4} θ + 3 s i n^{2} θ + 6}}{1 - c o s 2 θ} d θ$ (where c is a constant of integration)