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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2022 · Mathematics · Indefinite Integration

What is ∫ ( x)^ 1.5 -( x)^ 1.5 x · x d x equal to?

What is ∫(cos⁡x)1.5−(sin⁡x)1.5sin⁡x⋅cos⁡xdx\int \frac{(\cos x)^{1.5}-(\sin x)^{1.5}}{\sqrt{\sin x \cdot \cos x}} d x equal to?

Options

  1. Asin⁡x−cos⁡x+c\sqrt{\sin x}-\sqrt{\cos x}+c
  2. Bsin⁡x+cos⁡x+c\sqrt{\sin x}+\sqrt{\cos x}+c
  3. C2sin⁡x+2cos⁡x+c2 \sqrt{\sin x}+2 \sqrt{\cos x}+c
  4. D12sin⁡x+12cos⁡x+c\frac{1}{2} \sqrt{\sin x}+\frac{1}{2} \sqrt{\cos x}+c

Correct answer

C. 2sin⁡x+2cos⁡x+c2 \sqrt{\sin x}+2 \sqrt{\cos x}+c

Step-by-step solution

I=∫cos⁡3/2x−sin⁡3/2xsin⁡1/2x⋅cos⁡1/2xdx∫(sin⁡−1/2x⋅cos⁡x−cos⁡−1/2x⋅sin⁡x)dx=2⋅sin⁡1/2x+2⋅cos⁡1/2x+C=2sin⁡x+2cos⁡x+C\begin{aligned} & I=\int \frac{\cos ^{3 / 2} x-\sin ^{3 / 2} x}{\sin ^{1 / 2} x \cdot \cos ^{1 / 2} x} d x \\ & \int\left(\sin ^{-1 / 2} x \cdot \cos x-\cos ^{-1 / 2} x \cdot \sin x\right) d x \\ & =2 \cdot \sin ^{1 / 2} x+2 \cdot \cos ^{1 / 2} x+C=2 \sqrt{\sin x}+2 \sqrt{\cos x}+C\end{aligned}
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