If f(x)={ ² π x}{1+π^{x}}, then (f(x)+f(-x)) d x is equal to

Correct Answer is : 12x−sin⁡2πx4π+cr1/2 x- 2 π x/4 π+c_{r}21x−4πsin2πx+cr (where c is a constant of integration)

x/2- π x/2 π+c, (where c is a constant of integration)

1/2 x- 2 π x/4 π+c_{r} (where c is a constant of integration)

x/2- π x/2 π+c, (where c is a constant of integration)

{1}{1+π^{x}}+{ ² π x}{2 π}+c, (where c is a constant of integration)

Correct Answer is : 12x−sin⁡2πx4π+cr1/2 x- 2 π x/4 π+c_{r}21x−4πsin2πx+cr (where c is a constant of integration)

Test Index

MHTCET 16 May 2024 Shift 1 Paper

Question : 139 of 150

Marks: +1, -0

f(x)=\frac{\sin ^{2} \pi x}{1+\pi^{x}}

\int(f(x)+f(-x)) d x

is equal to

$\frac{x}{2}-\frac{\sin \pi x}{2 \pi}+c$ , (where c is a constant of integration)
$\frac{1}{2} x-\frac{\sin 2 \pi x}{4 \pi}+c_{r}$ (where c is a constant of integration)
$\frac{x}{2}-\frac{\cos \pi x}{2 \pi}+c$ , (where c is a constant of integration)
$\frac{1}{1+\pi^{x}}+\frac{\cos ^{2} \pi x}{2 \pi}+c$ , (where c is a constant of integration)

Solution:

Concept:

Use the property that

f(x)+f(-x)

simplifies by rewriting

f(-x)

with a common denominator, then integrate the resulting standard form.

Explanation:

Let

f(x)=\frac{\sin^2(\pi x)}{1+\pi^x}

Then

f(-x)=\frac{\sin^2(-\pi x)}{1+\pi^{-x}}=\frac{\sin^2(\pi x)}{1+\pi^{-x}}

Multiply the numerator and denominator of

f(-x)

\pi^x

f(-x)=\frac{\pi^x\sin^2(\pi x)}{1+\pi^x}

Add the two terms:

f(x)+f(-x)=\frac{\sin^2(\pi x)}{1+\pi^x}+\frac{\pi^x\sin^2(\pi x)}{1+\pi^x}=\sin^2(\pi x)

Therefore,

\int(f(x)+f(-x))dx=\int\sin^2(\pi x)dx

Use the identity

\sin^2\theta=\frac{1-\cos 2\theta}{2}

\int\sin^2(\pi x)dx=\int\frac{1-\cos(2\pi x)}{2}dx

Integrate term by term:

=\frac{x}{2}-\frac{\sin(2\pi x)}{4\pi}+c

This matches option B.

Answer:

Option B:

\frac{x}{2}-\frac{\sin(2\pi x)}{4\pi}+c

, where

c

is the constant of integration.

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