Integral Calculus – Definite Integration – JEE Main 29 Jan 2025 Shift 1

Solution:

Let $I = 80\int_{0}^{\frac{\pi}{4}}\frac{\sin \theta+\cos \theta}{9+16\sin 2\theta}d\theta$.

We manipulate the denominator to use the substitution $t = \sin \theta – \cos \theta$.

Recall $(\sin \theta – \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta – 2\sin \theta \cos \theta = 1 – \sin 2\theta$.

So, $\sin 2\theta = 1 – (\sin \theta – \cos \theta)^2$.

Substitute this into the denominator:

$9 + 16\sin 2\theta = 9 + 16(1 – (\sin \theta – \cos \theta)^2) = 9 + 16 – 16(\sin \theta – \cos \theta)^2 = 25 – 16(\sin \theta – \cos \theta)^2$.

Now, put $t = \sin \theta – \cos \theta$.

Then $dt = (\cos \theta + \sin \theta)d\theta$.

Change limits:

At $\theta = 0$, $t = 0 – 1 = -1$.

At $\theta = \frac{\pi}{4}$, $t = \frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} = 0$.

The integral becomes:

$I = 80 \int_{-1}^{0} \frac{dt}{25 – 16t^2}$.

$I = \frac{80}{16} \int_{-1}^{0} \frac{dt}{\frac{25}{16} – t^2} = 5 \int_{-1}^{0} \frac{dt}{(\frac{5}{4})^2 – t^2}$.

Using the formula $\int \frac{dx}{a^2 – x^2} = \frac{1}{2a} \ln |\frac{a+x}{a-x}|$:

$I = 5 \left[ \frac{1}{2(5/4)} \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} – t} \right| \right]_{-1}^{0}$.

$I = 5 \cdot \frac{2}{5} \left[ \ln \left| \frac{5+4t}{5-4t} \right| \right]_{-1}^{0}$.

$I = 2 \left[ \ln(1) – \ln \left| \frac{5-4}{5+4} \right| \right]$.

$I = 2 [0 – \ln(\frac{1}{9})] = 2 \ln 9 = 2 \ln(3^2) = 4 \ln 3$.

Ans. (3)