The integral 80 ∫_0^{π/4} (sin θ + cos θ) / (9 + 16 sin 2θ) dθ is equal to :
(1) 3 log_e 4
(2) 4 log

Q: Why did we choose $t = \sin \theta - \cos \theta$ instead of $\sin \theta + \cos \theta$?

Great question! We look for a substitution where the derivative is already present in the integral. The derivative of $\sin \theta - \cos \theta$ is $\cos \theta - (-\sin \theta) = \cos \theta + \sin \theta$, which is exactly our numerator. If we had chosen the other one, the derivative would be $\cos \theta - \sin \theta$, which doesn't match our numerator. Always look for the 'hidden' derivative!

Q: How do I know when to square the substitution variable?

Whenever you have a term like $\sin 2\theta$ in the denominator and you are using a substitution involving $\sin \theta$ and $\cos \theta$, squaring is the standard bridge. It utilizes the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ to isolate the double angle term. It is a classic JEE trick to convert trigonometric functions into algebraic ones.

Q: I got a negative value inside the logarithm. Is that allowed?

Remember that the formula involves the absolute value: $\ln |\frac{a+x}{a-x}|$. Even if the argument inside the absolute value is negative, the modulus makes it positive, which is valid for the natural logarithm. Always keep track of your signs, especially when evaluating the lower limit!

Question

The integral $80\int_{0}^{\frac{\pi}{4}}(\frac{\sin~	heta+\cos~	heta}{9+16~\sin~2	heta})d	heta$ is equal to :

Options:

(A) $3~\log_{e}4$

(B) $4~\log_{e}3$

(C) $6~\log_{e}4$

(D) $2~\log_{e}3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a trigonometric nightmare.

You see an integral like:

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

Your instinct might be to panic, but in the world of JEE Advanced, complexity is often just a mask for elegance. Our job is to peel back that mask.

### The Siren Song of the Numerator

Look at the numerator: $\sin 	heta + \cos 	heta$. In calculus, whenever you see a sum of sine and cosine, your intuition should immediately scream, "Substitution!"

This expression is the derivative of $\sin 	heta - \cos 	heta$. By defining $t = \sin 	heta - \cos 	heta$, we are transforming the entire landscape of the problem.

When we differentiate this, we get $dt = (\cos 	heta + \sin 	heta) d	heta$. Just like that, the entire numerator is absorbed into our differential $dt$.

### The Algebraic Bridge

Now, we face the denominator: $9 + 16 \sin 2	heta$. We need to convert the trigonometric $\sin 2	heta$ into an algebraic expression involving $t$.

We square our substitution: $t^2 = (\sin 	heta - \cos 	heta)^2$. Expanding this, we get:

$$t^2 = \sin^2 	heta + \cos^2 	heta - 2 \sin 	heta \cos 	heta$$

Using the identity $\sin^2 	heta + \cos^2 	heta = 1$ and the double angle formula $2 \sin 	heta \cos 	heta = \sin 2	heta$, we arrive at $t^2 = 1 - \sin 2	heta$. Rearranging this gives us $\sin 2	heta = 1 - t^2$.

### The Transformation of Limits

Never forget your limits! When we change the variable, the boundaries of our world change too.

At $	heta = 0$, $t = \sin(0) - \cos(0) = -1$. At $	heta = \frac{\pi}{4}$, $t = \sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 0$.

Our integral now spans from $-1$ to $0$:

$$I = 80 \int_{-1}^{0} \frac{dt}{9 + 16(1 - t^2)}$$

Simplifying the denominator, we get $9 + 16 - 16t^2 = 25 - 16t^2$. Factoring out $16$, we obtain:

$$I = 5 \int_{-1}^{0} \frac{dt}{(\frac{5}{4})^2 - t^2}$$

### The Final Victory

This is a standard integral of the form:

$$\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} ight|$$

With $a = \frac{5}{4}$, the constant becomes $5 	imes \frac{1}{2(5/4)} = 2$. Evaluating the limits:

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

The upper limit gives us $\ln(1) = 0$. The lower limit gives us $-2 \ln \left| \frac{5/4 - 1}{5/4 + 1} ight| = -2 \ln \left( \frac{1/4}{9/4} ight) = -2 \ln \left( \frac{1}{9} ight)$.

Since $\ln(1/9) = \ln(3^{-2}) = -2 \ln 3$, the final result is:

**$I = 4 \ln 3$**

Animated Solution for Mathematics - Definite Integration: The integral $80 \int_{0}^{\frac{π}{4}} (\frac{s i n θ + c o s θ}{9 + 16 s i n 2 θ}) d θ$ is equal to :

Select Answer:

Visualized Solution

Analyze the Integral Structure

The Magic Substitution

Differentiating the Substitution

Relating $sin 2 θ$ to $t$

Expanding the Squared Term

Isolating $sin 2 θ$

Transforming the Limits

Constructing the New Integral

Simplifying the Denominator

Applying the Standard Integral Formula

Substituting into the Formula

Evaluating the Limits

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Siren Song of the Numerator

The Algebraic Bridge

The Transformation of Limits

The Final Victory

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral 80∫04π​​(9+16 sin 2θsin θ+cos θ​)dθ is equal to :