New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Definite Integration question

2002 · Shift 0 · Q82

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Definite Integration
2002 · Shift 0 · Q82

Definite Integration question

2002 · Shift 0 · Q82

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1

\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx

is

A
${{{\pi ^2}} \over 4}$
B
${{\pi ^2}}$
C
zero
D
${\pi \over 2}$

View written solutionFree

Correct answer: B

Let $I=\int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x}\,dx.$
Split the integrand: $I=\int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx+\int_{-\pi}^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx.$ So write $I=I_1+I_2.$
Check parity of each part.

For $f_1(x)=\frac{2x}{1+\cos^2 x},$ we have $x$ odd and $1+\cos^2 x$ even, so $f_1(x)$ is odd. Hence, $I_1=\int_{-\pi}^{\pi} f_1(x)\,dx=0.$
For $f_2(x)=\frac{2x\sin x}{1+\cos^2 x},$ $x$ is odd, $\sin x$ is odd, so $x\sin x$ is even; denominator is even. Thus $f_2(x)$ is even. Therefore, $I_2=2\int_0^{\pi} \frac{2x\sin x}{1+\cos^2 x}\,dx=4\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx.$ So, $I=4\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx.$

Now use the property with substitution $x\mapsto \pi-x$ . Let $J=\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx.$ Put $x=\pi-t$ . Then $dx=-dt$ , and as $x:0\to\pi$ , $t:\pi\to 0$ . Hence $J=\int_0^{\pi} \frac{(\pi-t)\sin t}{1+\cos^2 t}\,dt.$
Add the two expressions for $J$ :

=\pi\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx.$$ Thus, $$J=\frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx.$$ 6. Evaluate the remaining integral: $$\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx.$$ Let $$u=\cos x,\qquad du=-\sin x\,dx.$$ When $x=0$, $u=1$; when $x=\pi$, $u=-1$. Therefore, $$\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx =\int_1^{-1} \frac{-du}{1+u^2} =\int_{-1}^{1} \frac{du}{1+u^2}.$$ Now, $$\int \frac{du}{1+u^2}=\tan^{-1}u,$$ so $$\int_{-1}^{1} \frac{du}{1+u^2}=\tan^{-1}(1)-\tan^{-1}(-1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}.$$ Hence, $$J=\frac{\pi}{2}\cdot \frac{\pi}{2}=\frac{\pi^2}{4}.$$ 7. Therefore, $$I=4J=4\cdot \frac{\pi^2}{4}=\pi^2.$$ 8. So the value of the integral is $$\boxed{\pi^2}.$$ This corresponds to option **B**.

More from Definite Integration

Let [.] denote the greatest integer function. If $0 \int e^{3} [\frac{1}{e ^{x - 1}}] d x = α - lo g_{e} 2$ , then $α^{3}$ is equal to .2025 · Numerical
Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10 \int_{1}^{1} f (t) dt = 5 x f (x) - x^{5} - 9$ for all $x ⩾ 1$ , then the value of $f (3)$ is :2025 · MCQ
Let $(a, b)$ be the point of intersection of the curve $x^{2} = 2 y$ and the straight line $y - 2 x - 6 = 0$ in the second quadrant. Then the integral $I = \int_{a}^{b} \frac{9 x ^{2}}{1 + 5 ^{x}} d x$ is equal to :2025 · MCQ
$4 \int_{0}^{1} (\frac{1}{3 + x ^{2} + 1 + x ^{2}}) d x - 3 lo g_{e} (3)$ is equal to :2025 · MCQ
Let the domain of the function $f (x) = lo g_{2} lo g_{4} lo g_{6} (3 + 4 x - x^{2})$ be $(a, b)$ . If $\int_{0}^{b - a} [x^{2}] d x = p - q - r, p, q, r \in N, gcd (p, q, r) = 1$ , where $[\cdot]$ is the…2025 · MCQ
The integral $\int_{0}^{π} \frac{8 x d x}{4 c o s ^{2} x + s i n ^{2} x}$ is equal to2025 · MCQ
The value of $\int_{- 1}^{1} \frac{( 1 + ∣ x ∣ - x ) e ^{x} + ( ∣ x ∣ - x ) e ^{- x}}{e ^{x} + e ^{- x}} d x$ is equal to2025 · MCQ
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is :2025 · MCQ