NDA 2026 — Mathematics PYQ

1. Analyze the Given Integral $k$

We are given:

$$k=\frac{1}{a+b}{\int }_0^{\frac{\pi }{2}}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx\dots \text{(Eq. 1)}$$

From this, we can express the integral as:

$${\int }_0^{\frac{\pi }{2}}\frac{a\sin x+b\cos x}{\sin x+\cos x}dx=k(a+b)\dots \text{(Eq. 2)}$$

2. Apply Definite Integral Property

Let the integral we want to evaluate be $I$:

$$I={\int }_0^{\frac{\pi }{2}}\frac{a\cos x+b\sin x}{\sin x+\cos x}dx$$

Using the property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$ with $a=\frac{\pi }{2}$:

$$I={\int }_0^{\frac{\pi }{2}}\frac{a\cos (\frac{\pi }{2}-x)+b\sin (\frac{\pi }{2}-x)}{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}dx$$

Since $\cos (\frac{\pi }{2}-x)=\sin x$ and $\sin (\frac{\pi }{2}-x)=\cos x$:

$$I={\int }_0^{\frac{\pi }{2}}\frac{a\sin x+b\cos x}{\cos x+\sin x}dx\dots \text{(Eq. 3)}$$

3. Final Conclusion

Comparing (Eq. 3) with (Eq. 2), we can see that:

$$I=k(a+b)$$

Correct Option: (c)