Animated Solution for Mathematics - Definite Integration: The integral ${\int }_0^{\pi /4}\frac{136\sin x}{3\sin x+5\cos x}dx$ is equal to :

Welcome, future engineer. Today, we are going to dissect a problem that is a staple of the JEE Advanced curriculum. It is a problem that tests not just your ability to integrate, but your ability to see the hidden structure within a seemingly chaotic expression.

We are looking at the integral:

At first glance, it looks like a nightmare. You have a trigonometric function in the numerator and a different one in the denominator. There is no obvious substitution.

But here is the secret: in the world of JEE mathematics, whenever you see a linear combination of sine and cosine in both the numerator and the denominator, you are looking at a specific archetype. It is a puzzle waiting to be solved by the method of undetermined coefficients.

Imagine you are a surgeon. You need to operate on this fraction to make it manageable. We want to rewrite the numerator, $136\sin x$, as a combination of the denominator, $f(x)=3\sin x+5\cos x$, and its derivative, $f^{\prime }(x)=3\cos x-5\sin x$.

Why? Because if we can do that, the integral splits into two parts: one where the denominator cancels out, and one where the numerator is exactly the derivative of the denominator. The latter is the golden ticket to a natural logarithm.

So, we set up the equation:

Here, $A$ and $B$ are our unknown constants.

Now, we expand and group the terms. On the right side, we have $(3A-5B)\sin x+(5A+3B)\cos x$. On the left, we have $136\sin x+0\cos x$.

By comparing the coefficients, we get a system of two linear equations:

Solving this is straightforward but requires precision. From the second equation, $B=-\frac{5}{3}A$. Substituting this into the first, we get:

This leads to $\frac{34}{3}A=136$, giving us $A=12$. Consequently, $B=-20$. We have successfully decomposed the numerator!

Now, the integral becomes:

Splitting this into two integrals, we get:

The first part is trivial: $12x$. The second part is the classic $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|$. So, our integral is $12x-20\ln |3\sin x+5\cos x|$ evaluated from $0$ to $\pi /4$.

Evaluating at the upper limit $\pi /4$, we get $12(\frac{\pi }{4})-20\ln |3\sin (\frac{\pi }{4})+5\cos (\frac{\pi }{4})|$. Since $\sin (\frac{\pi }{4})=\cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$, this becomes:

At the lower limit $0$, we get $12(0)-20\ln |3(0)+5(1)|=-20\ln 5$. Subtracting the lower limit from the upper limit, we get $3\pi -20\ln (4\sqrt{2})+20\ln 5$.

Using the properties of logarithms, $4\sqrt{2}=2^2\cdot 2^{1/2}=2^{5/2}$. Thus, $20\ln (2^{5/2})=20\cdot \frac{5}{2}\ln 2=50\ln 2$.

Our final answer is $3\pi -50\ln 2+20\ln 5$. You see? It was not a nightmare; it was a beautiful, logical progression. Keep this method in your toolkit, and no integral will ever intimidate you again.