Evaluate: ∫ (0 to π/4) (sin x + cos x) / (9 + 16 sin 2x) dx.

Q: Wait, why do we choose $t = \sin x - \cos x$ instead of $t = \sin x + \cos x$?

Great question! If we chose $t = \sin x + \cos x$, the derivative would be $\cos x - \sin x$, which doesn't match our numerator. By choosing $t = \sin x - \cos x$, the derivative is $\cos x + \sin x$, which is exactly what we have in the numerator. It is all about making the substitution work for the differential $dx$!

Q: Why did the limits of integration change from $[0, \pi/4]$ to $[-1, 0]$?

When we perform a substitution, we are changing the variable of integration from $x$ to $t$. The original limits $0$ and $\pi/4$ are values for $x$. We must find the corresponding values for $t$ using our substitution $t = \sin x - \cos x$. At $x=0$, $t = \sin 0 - \cos 0 = -1$. At $x=\pi/4$, $t = \sin(\pi/4) - \cos(\pi/4) = 0$. Always remember to update your limits!

Q: How do I know when to use the standard integral formula $\int \frac{dx}{a^2 - x^2}$?

Whenever you encounter an integral where the denominator is a quadratic expression in terms of your variable (like $25 - 16t^2$), you should aim to manipulate it into a standard form. By factoring out the coefficient of $t^2$, you can easily identify the constant $a^2$ and use the standard logarithmic integral formula. It is a powerful tool in your JEE toolkit!

Question

Evaluate: $\int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to demystify a seemingly complex definite integral:

$$I = \int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx$$

At first glance, this function might look like a labyrinth, but there is a clear, elegant path through it. Geometrically, this integral represents the area under the curve from $x = 0$ to $x = \frac{\pi}{4}$.

### The Numerator's Secret

In calculus, the numerator is often a signpost. Look at $\sin x + \cos x$; it is the derivative of $\sin x - \cos x$.

This is our 'Aha!' moment. Let us define our substitution variable as $t = \sin x - \cos x$.

When we differentiate this with respect to $x$, we get $dt = (\cos x + \sin x) dx$. Just like that, the entire numerator is accounted for!

### The Trigonometric Identity Trick

Now, we must address the denominator $9 + 16 \sin 2x$. We need to express $\sin 2x$ in terms of $t$.

Let us take our substitution $t = \sin x - \cos x$ and square it:

$$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x$$

Using the fundamental identity $\sin^2 x + \cos^2 x = 1$ and the double angle identity $\sin 2x = 2 \sin x \cos x$, this simplifies to $t^2 = 1 - \sin 2x$. Rearranging, we find that $\sin 2x = 1 - t^2$.

### The Transformation of Limits

Before we proceed, we must be careful. When we change the variable from $x$ to $t$, the boundaries of our integral must also change.

For the lower limit, when $x = 0$:
$t = \sin 0 - \cos 0 = -1$.

For the upper limit, when $x = \frac{\pi}{4}$:
$t = \sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 0$.

Our integral now spans from $t = -1$ to $t = 0$.

### The Final Stretch

Substituting everything back into our integral, we get:

$$I = \int_{-1}^0 \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^0 \frac{dt}{25 - 16t^2}$$

To solve this, we factor out $16$ to match the standard form $\int \frac{dx}{a^2 - x^2}$:

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

Using the standard formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right|$, with $a = \frac{5}{4}$, we obtain:

$$I = \frac{1}{16} \left[ \frac{1}{2(5/4)} \ln \left| \frac{5/4 + t}{5/4 - t} \right| \right]_{-1}^0 = \frac{1}{40} \left[ \ln \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^0$$

Evaluating at the limits:

$$I = \frac{1}{40} [\ln(1) - \ln(1/9)] = \frac{1}{40} \ln(9) = \frac{1}{20} \ln(3)$$

You have done it! Through careful substitution and identity manipulation, we have arrived at the elegant result of **$\frac{1}{20} \ln 3$**.

Animated Solution for Mathematics - Definite Integration: Evaluate: $\int_{0}^{π /4} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x$ .

Enter Numerical Value:

Visualized Solution

The Definite Integral

Identifying the Substitution

Expressing $sin 2 x$ in terms of $t$

Changing the Limits of Integration

Substituting into the Integral

Factoring to Standard Form

Using the Standard Formula

Evaluating Upper and Lower Limits

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Numerator's Secret

The Trigonometric Identity Trick

The Transformation of Limits

The Final Stretch

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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; } Evaluate: ∫0π/4​9+16sin2xsinx+cosx​dx.