The value of ∫ (1 - cos θ) ^ (3)/(10) (1 + cos θ) ^ (13)/(10) dθ is equal to (where, c & is the constant of integration)

Correct answer: B. (5)/(8)(tan (θ)/(2))(8)/(5) + c ∫ ((1 - cos θ)(3)/(10))/((1 + cos θ)(13)/(10))dθ = ∫ (2(3)/(10)(sin (θ)/(2))(3)/(5))/(2(13)/(10)(cos (θ)/(2))(13)/(5))dθ = (1)/(2) ∫ (tan (θ)/(2))(3)/(5)sec^ 2 (θ)/(2)dθ Let, tan (θ)/(2) = t ⇒ (1)/(2)sec^ 2 (θ)/(2)dθ = dt = ∫ t(3)/(5)dt = t(8)/(5)((8)/(5)) + c = (5)/(8)(tan (θ)/(2))(8)/(5) + c

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of ∫ (1 - cos θ) ^ (3)/(10) (1 + cos θ) ^ (13)/(10) dθ is...

The value of

∫ \dfrac{{\left(1 - cos \theta\right)}^{\dfrac{3}{10}}}{{\left(1 + cos \theta\right)}^{\dfrac{13}{10}}}d\theta

is equal to (where,

c

& is the constant of integration)

Options

A $\dfrac{5}{8}{\left(tan \theta\right)}^{\dfrac{5}{8}} + c$
B $\dfrac{5}{8}\left(tan \dfrac{θ}{2}\right)\dfrac{8}{5} + c$
C $\dfrac{5}{16}\left(tan \dfrac{θ}{2}\right)\dfrac{8}{5} + c$
D $\dfrac{5}{8}{\left(tan \theta\right)}^{\dfrac{5}{16}} + c$

Correct answer

B. $\dfrac{5}{8}\left(tan \dfrac{θ}{2}\right)\dfrac{8}{5} + c$

Step-by-step solution

∫ \dfrac{\left(1 - cos \theta\right)\dfrac{3}{10}}{\left(1 + cos \theta\right)\dfrac{13}{10}}d\theta = ∫ \dfrac{2\dfrac{3}{10}\left(sin \dfrac{θ}{2}\right)\dfrac{3}{5}}{2\dfrac{13}{10}\left(cos \dfrac{θ}{2}\right)\dfrac{13}{5}}d\theta

= \dfrac{1}{2} ∫ \left(tan \dfrac{θ}{2}\right)\dfrac{3}{5}sec^{2}\dfrac{θ}{2}d\theta

Let,

tan \dfrac{θ}{2} = t ⇒ \dfrac{1}{2}sec^{2}\dfrac{θ}{2}d\theta = dt

= ∫ t\dfrac{3}{5}dt = t\dfrac{8}{5}\left(\dfrac{8}{5}\right) + c

= \dfrac{5}{8}\left(tan \dfrac{θ}{2}\right)\dfrac{8}{5} + c

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