The value of ∫ sin^ 3 x cos x dx is equal to (where, c & is the constant of integration)

Correct answer: B. - (2)/(3) (cos x) ^ (3)/(2) + (2)/(7) ( x) ^ (7)/(2) + c ∫ sin^ 2 x cos x sin xdx Substituting, cos x = t and sin xdx = - dt = - ∫ (1 - t^ 2 ) t dt = - ∫ (t(1)/(2) - t(5)/(2))dt = (- t(3)/(2))/(32) + (t(7)/(2))/(72) + c = (- 2)/(3)(cos x)(3)/(2) + (2)/(7)(cos x)(7)/(2) + c

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of ∫ sin^ 3 x cos x dx is equal to (where, c & is the...

The value of

∫ sin^{3}x\sqrt{cos x}dx

is equal to (where,

c

& is the constant of integration)

Options

A $\dfrac{2}{3}{\left(cos x\right)}^{\dfrac{3}{2}} + \dfrac{2}{7}{\left(cos x\right)}^{\dfrac{7}{2}} + c$
B $- \dfrac{2}{3}{\left(cos x\right)}^{\dfrac{3}{2}} + \dfrac{2}{7}{\left(\cos x\right)}^{\dfrac{7}{2}} + c$
C $\left(- \dfrac{2}{3}\right){\left(cos x\right)}^{3} + \dfrac{2}{5}{\left(cos x\right)}^{\dfrac{5}{2}} + c$
D $\dfrac{3}{2}{\left(cos x\right)}^{\dfrac{3}{2}} + \dfrac{5}{2}{\left(cos x\right)}^{\dfrac{7}{2}} + c$

Correct answer

B. $- \dfrac{2}{3}{\left(cos x\right)}^{\dfrac{3}{2}} + \dfrac{2}{7}{\left(\cos x\right)}^{\dfrac{7}{2}} + c$

Step-by-step solution

∫ sin^{2}x\sqrt{cos x}sin xdx

Substituting,

cos x = t

and

sin xdx = - dt

= - ∫ \left(1 - t^{2}\right)\sqrt{t}dt

= - ∫ \left(t\dfrac{1}{2} - t\dfrac{5}{2}\right)dt

= \dfrac{- t\dfrac{3}{2}}{32} + \dfrac{t\dfrac{7}{2}}{72} + c

= \dfrac{- 2}{3}\left(cos x\right)\dfrac{3}{2} + \dfrac{2}{7}\left(cos x\right)\dfrac{7}{2} + c

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