Q112 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

$\int \sqrt{4 \cos ^2 x-5 \sin ^2 x} \cos x d x=$

  • A
    $\frac{1}{2} \cos x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+c$
  • B
    $\frac{1}{2} \sin x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \cos ^{-1}\left(\frac{3 \cos x}{2}\right)+c$
  • C
    $\frac{1}{2} \cos x \sqrt{1-9 \cos ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \cos x}{2}\right)+c$
  • Correct answer
    D
    $\frac{1}{2} \sin x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+c$

Solution

$\begin{aligned} & \text { } I=\int \sqrt{4 \cos ^2 x-5 \sin ^2 x} \cos x d x \\ & =\int \sqrt{4-9 \sin ^2 x} \cos x d x ; \text { Let } \sin x=\frac{2}{3} t \Rightarrow d x=\frac{2 d t}{3 \cos x} \\ & I=\int \frac{4}{3} \sqrt{1-t^2} d t \Rightarrow I=\frac{4}{3}\left[\frac{t}{2} \sqrt{1-t^2}+\frac{1}{2} \sin ^{-1} t\right]+C \\ & \Rightarrow I=\frac{4}{3}\left[\frac{3}{4} \sin x \sqrt{1-\frac{9}{4} \sin ^2 x}+\frac{1}{2} \sin ^{-1}\left(\frac{3}{2} \sin x\right)\right]+C \\ & \Rightarrow I=\frac{1}{2} \sin x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+C\end{aligned}$

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

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