The following integral ∫_(π/4)^(π/2) (2 csc x)¹⁷ dx is equal to (a) ∫₀^(ln(1+√2)) 2(eᵘ + e⁻ᵘ)¹⁶ du (

Q: Why did we choose $\csc x + \cot x = e^u$? It feels like magic.

It is a classic JEE trick! This substitution is specifically designed to linearize the cosecant function. By mapping the trigonometric domain to the exponential domain, we transform a complex power into a simple binomial expansion. It is not magic; it is the art of choosing the right coordinate system for your problem.

Q: Why do we split the power 17 into 16 and 1?

We need one $\csc x$ term to pair with the $dx$ to form the differential $du$. By isolating one factor, we create the perfect setup for the substitution $\csc x dx = -du$. Without this split, we would have no way to account for the differential $dx$ in terms of $du$.

Q: What happens to the negative sign when we flip the limits?

The property $\int_a^b f(u) du = -\int_b^a f(u) du$ is our best friend here. Since our differential substitution gave us a $-du$, flipping the limits from $[\ln(1+\sqrt{2}), 0]$ to $[0, \ln(1+\sqrt{2})]$ absorbs the negative sign, leaving us with a clean, positive integral.

Question

The following integral $\int_{\pi/4}^{\pi/2} (2 \csc x)^{17} dx$ is equal to

Options:

(A) $\int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$

(B) $\int_{0}^{\ln(1+\sqrt{2})} (e^u + e^{-u})^{17} du$

(C) $\int_{0}^{\ln(1+\sqrt{2})} (e^u - e^{-u})^{17} du$

(D) $\int_{0}^{\ln(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The integral provided is $I = \int_{\pi/4}^{\pi/2} (2 \csc x)^{17} dx$. The exponent $17$ is a classic 'scare tactic' designed to discourage direct integration by parts.

To solve this, we must employ a structural transformation. We define the substitution:
$$\csc x + \cot x = e^u$$

### The Differential Dance

We differentiate the substitution with respect to $x$. The derivative of $\csc x + \cot x$ is $-\csc x \cot x - \csc^2 x$, which factors into $-\csc x(\cot x + \csc x)$.

Substituting our original definition, we obtain:
$$-\csc x (e^u) dx = e^u du$$

The $e^u$ terms cancel out, leading to the elegant differential relation:
$$\csc x dx = -du$$

### The Identity Bridge

To express $2 \csc x$ in terms of $u$, we use the identity $\csc^2 x - \cot^2 x = 1$. Factoring this as a difference of squares, we have:
$$(\csc x - \cot x)(\csc x + \cot x) = 1$$

Since $\csc x + \cot x = e^u$, it follows that $\csc x - \cot x = e^{-u}$. Adding these two equations eliminates the cotangent terms:
$$2 \csc x = e^u + e^{-u}$$

### The Final Assembly

We rewrite the integrand as $(2 \csc x)^{16} \cdot (2 \csc x)$. Substituting our exponential expressions, we get:
$$(e^u + e^{-u})^{16} \cdot 2 \cdot (-du)$$

Next, we transform the limits of integration. At the lower limit $x = \pi/4$:
$$e^u = \csc(\pi/4) + \cot(\pi/4) = \sqrt{2} + 1 \Rightarrow u = \ln(1+\sqrt{2})$$

At the upper limit $x = \pi/2$:
$$e^u = \csc(\pi/2) + \cot(\pi/2) = 1 + 0 = 1 \Rightarrow u = 0$$

The integral becomes:
$$I = \int_{\ln(1+\sqrt{2})}^{0} (e^u + e^{-u})^{16} (-2 du)$$

By flipping the limits to absorb the negative sign, we arrive at the final, simplified form:
$$I = \int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$$

This result demonstrates the power of calculus in distilling chaotic trigonometric expressions into clean, solvable exponential forms.

Animated Solution for Mathematics - Definite Integration: The following integral $\int_{π /4}^{π /2} (2 csc x)^{17} d x$ is equal to

Select Answer:

Visualized Solution

The Challenge: High Powers of $csc x$

Differentiating the Substitution

Simplifying the Differential Expression

Finding the Conjugate Expression

Expressing $2 csc x$ in terms of $u$

Changing the Limits of Integration

Rewriting the Integral

Substituting the Values

Final Simplification and Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Differential Dance

The Identity Bridge

The Final Assembly

Similar Questions

Evaluate the following: $\int_{0}^{π /2} \frac{x s i n x c o s x}{c o s ^{4} x + s i n ^{4} x} d x$

Evaluate $\int_{0}^{π} e^{∣ c o s x ∣} (2 sin (\frac{1}{2} cos x) + 3 cos (\frac{1}{2} cos x)) sin x d x$ .

The value of $\int_{0}^{π} \frac{e ^{c o s x} s i n x}{( 1 + c o s ^{2} x ) ( e ^{c o s x} + e ^{- c o s x} )} d x$ is equal to

The value of the integral $\int_{- l n 2}^{l n 2} e^{x} (ln (e^{x} + 1 + e^{2 x})) d x$ is equal to

The value of $\int_{- π /2}^{π /2} (\frac{1 + s i n ^{2} x}{1 + π ^{s i n x}}) d x$ is

Integrate $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ .

Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

The value of the integral $\int_{- π /2}^{π /2} sin^{4} x (1 + lo g (\frac{2 + s i n x}{2 - s i n x})) d x$ is :

The value of $\int_{- π /2}^{π /2} \frac{x ^{2} c o s x}{1 + e ^{x}} d x$ is equal to

Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

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 following integral ∫π/4π/2​(2cscx)17dx is equal to