पिछले वर्ष के नीट प्रश्न - समाकलन गणित

2019 प्रश्न 1. मूल्यांकन कीजिए: $\int \frac{\sin x}{1+\cos x} dx$.

We can use the following substitution:

$u = \cos x$

$du = -\sin x dx$

So, the integral becomes:

$\int \frac{du}{1+u} = \ln(1+u) + C$

Substituting back in, we get:

$\ln(1+\cos x) + C$

2018 प्रश्न 2. मूल्यांकन कीजिए: $\int \frac{x^2}{(x^2+1)^2} dx$.

We can use the following substitution:

$u = x^2+1$

$du = 2x dx$

So, the integral becomes:

$\frac{1}{2} \int \frac{du}{u^2} = -\frac{1}{2u} + C$

Substituting back in, we get:

$-\frac{1}{2(x^2+1)} + C$