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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Definite Integration

NDA 2026 · Mathematics · Definite Integration

Passage: Let ∫_0^ (π)/(2) (a x+b x)/((a+b)( x+ x))\,dx=k ....

Passage: Let ∫0π2asin⁡x+bcos⁡x(a+b)(sin⁡x+cos⁡x) dx=k\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{a\sin x+b\cos x}{(a+b)(\sin x+\cos x)}\,dx=k.

Question: What is ∫0π2acos⁡x+bsin⁡xsin⁡x+cos⁡x dx\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\sin x+\cos x}\,dx equal to ?

Options

  1. Akk
  2. B2k2k
  3. Ck(a+b)k(a+b)
  4. Dka+b\dfrac{k}{a+b}

Correct answer

C. k(a+b)k(a+b)

Step-by-step solution

Given k=∫0π2asin⁡x+bcos⁡x(a+b)(sin⁡x+cos⁡x) dx\displaystyle k = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{(a+b)(\sin x+\cos x)}\,dx

Multiplying both sides by (a+b)(a+b):

k(a+b)=∫0π2asin⁡x+bcos⁡xsin⁡x+cos⁡x dx\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin x+b\cos x}{\sin x+\cos x}\,dx

Using the definite integral property ∫0af(x) dx=∫0af(a−x) dx\displaystyle \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx, replacing xx with π2−x\dfrac{\pi}{2}-x:

k(a+b)=∫0π2asin⁡(π2−x)+bcos⁡(π2−x)sin⁡(π2−x)+cos⁡(π2−x) dx\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\sin\left(\dfrac{\pi}{2}-x\right)+b\cos\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)}\,dx

k(a+b)=∫0π2acos⁡x+bsin⁡xcos⁡x+sin⁡x dx\displaystyle k(a+b) = \int_0^{\dfrac{\pi}{2}}\dfrac{a\cos x+b\sin x}{\cos x+\sin x}\,dx

This matches the required integral. Thus, the value is k(a+b)k(a+b).

Answer: k(a+b)k(a+b)
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