Q77 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics

$\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin x}\,dx =$

  • A
    $\log(\cos x) + \cos x + c$, where c is the constant of integration
  • Correct answer
    B
    $\log(\sin x) - \sin x + c$, where c is the constant of integration
  • C
    $\log(\sin x) + \sin x + c$, where c is the constant of integration
  • D
    $\log(\cos x) - \cos x + c$, where c is the constant of integration

Solution

Let $I = \displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin x} \,dx$

We can write $\cos^3 x = \cos^2 x \cos x = (1 - \sin^2 x) \cos x$

$I = \displaystyle\int \dfrac{1 - \sin^2 x}{\sin^2 x + \sin x} \cos x \,dx$

Let $\sin x = t$, then $\cos x \,dx = dt$

Substituting these into the integral, we get:

$I = \displaystyle\int \dfrac{1 - t^2}{t^2 + t} \,dt$

$I = \displaystyle\int \dfrac{(1 - t)(1 + t)}{t(t + 1)} \,dt$

$I = \displaystyle\int \dfrac{1 - t}{t} \,dt$

$I = \displaystyle\int \left(\dfrac{1}{t} - 1\right) \,dt$

$I = \log(t) - t + c$

Substituting back $t = \sin x$, we get:

$I = \log(\sin x) - \sin x + c$

Answer: $\log(\sin x) - \sin x + c$, where c is the constant of integration

Hence, the correct answer is option: 'B'.

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