The value of \int_\frac\pi3^\frac\pi2 \frac(2+3 \sin x)\sin x(1+\cos x) d x is equal to :

Definite Integration PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q38 with Solution

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Correct answer: A

We need to evaluate $I=\int_{\pi/3}^{\pi/2} \frac{2+3\sin x}{\sin x(1+\cos x)}\,dx.$
Split the integrand: $\frac{2+3\sin x}{\sin x(1+\cos x)}=\frac{2}{\sin x(1+\cos x)}+\frac{3\sin x}{\sin x(1+\cos x)}.$ So, $I=\int_{\pi/3}^{\pi/2} \frac{2}{\sin x(1+\cos x)}dx+3\int_{\pi/3}^{\pi/2}\frac{1}{1+\cos x}dx.$
Simplify the first term using $\frac{1}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin^3 x}?$ But a better identity is: $\frac{1}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin x(1-\cos^2 x)}=\frac{1-\cos x}{\sin^3 x},$ which is not the easiest to integrate directly.

Instead, use half-angle identities: $1+\cos x=2\cos^2\frac x2,\qquad \sin x=2\sin\frac x2\cos\frac x2.$ Hence, $\sin x(1+\cos x)=4\sin\frac x2\cos^3\frac x2,$ and the integrand can be handled more simply by rewriting the two parts separately.

For the second part, $\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}=\frac12\sec^2\frac x2.$ Therefore, $3\int \frac{1}{1+\cos x}dx=3\int \frac12\sec^2\frac x2\,dx=3\tan\frac x2.$
For the first part, $\frac{2}{\sin x(1+\cos x)}.$ Using $\sin x=2\sin\frac x2\cos\frac x2,\quad 1+\cos x=2\cos^2\frac x2,$ we get $\frac{2}{\sin x(1+\cos x)}=\frac{2}{4\sin(x/2)\cos^3(x/2)}=\frac{1}{2\sin(x/2)\cos^3(x/2)}.$ Now write in terms of $t=\tan\frac x2$ . Then $dx=\frac{2}{1+t^2}dt,\quad \sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad 1+\cos x=\frac{2}{1+t^2}.$ Thus, $\sin x(1+\cos x)=\frac{2t}{1+t^2}\cdot \frac{2}{1+t^2}=\frac{4t}{(1+t^2)^2}.$ So,

=\frac{2+\frac{6t}{1+t^2}}{\frac{4t}{(1+t^2)^2}}\cdot \frac{2}{1+t^2}dt.$$ Simplify: $$=\frac{2(1+t^2)+6t}{1+t^2}\cdot \frac{(1+t^2)^2}{4t}\cdot \frac{2}{1+t^2}dt$$ $$=\frac{2(1+t^2+3t)}{1+t^2}\cdot \frac{1+t^2}{2t}dt$$ $$=\frac{1+t^2+3t}{t}dt$$ $$=\left(t+3+\frac{1}{t}\right)dt.$$ 6. Change the limits: - When $x=\pi/3$, $$t=\tan\frac{\pi}{6}=\frac{1}{\sqrt3}.$$ - When $x=\pi/2$, $$t=\tan\frac{\pi}{4}=1.$$ Therefore, $$I=\int_{1/\sqrt3}^{1}\left(t+3+\frac1t\right)dt.$$ 7. Integrate: $$I=\left[\frac{t^2}{2}+3t+\ln t\right]_{1/\sqrt3}^{1}.$$ Now evaluate: At $t=1$: $$\frac12+3+\ln 1=\frac72.$$ At $t=\frac{1}{\sqrt3}$: $$\frac{1}{2\cdot 3}+\frac{3}{\sqrt3}+\ln\left(\frac{1}{\sqrt3}\right) =\frac16+\sqrt3-\ln \sqrt3.$$ Hence, $$I=\frac72-\left(\frac16+\sqrt3-\ln\sqrt3\right)$$ $$=\frac{21-1}{6}-\sqrt3+\ln\sqrt3$$ $$=\frac{20}{6}-\sqrt3+\ln\sqrt3$$ $$=\frac{10}{3}-\sqrt3+\ln\sqrt3.$$ 8. Compare with options: $$\boxed{\frac{10}{3}-\sqrt3+\log_e\sqrt3}$$ This matches **Option A**.

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View written solutionFree

Correct answer: A

We need to evaluate $I=\int_{\pi/3}^{\pi/2} \frac{2+3\sin x}{\sin x(1+\cos x)}\,dx.$
Split the integrand: $\frac{2+3\sin x}{\sin x(1+\cos x)}=\frac{2}{\sin x(1+\cos x)}+\frac{3\sin x}{\sin x(1+\cos x)}.$ So, $I=\int_{\pi/3}^{\pi/2} \frac{2}{\sin x(1+\cos x)}dx+3\int_{\pi/3}^{\pi/2}\frac{1}{1+\cos x}dx.$
Simplify the first term using $\frac{1}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin^3 x}?$ But a better identity is: $\frac{1}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin x(1-\cos^2 x)}=\frac{1-\cos x}{\sin^3 x},$ which is not the easiest to integrate directly.

For the second part, $\frac{1}{1+\cos x}=\frac{1}{2\cos^2(x/2)}=\frac12\sec^2\frac x2.$ Therefore, $3\int \frac{1}{1+\cos x}dx=3\int \frac12\sec^2\frac x2\,dx=3\tan\frac x2.$
For the first part, $\frac{2}{\sin x(1+\cos x)}.$ Using $\sin x=2\sin\frac x2\cos\frac x2,\quad 1+\cos x=2\cos^2\frac x2,$ we get $\frac{2}{\sin x(1+\cos x)}=\frac{2}{4\sin(x/2)\cos^3(x/2)}=\frac{1}{2\sin(x/2)\cos^3(x/2)}.$ Now write in terms of $t=\tan\frac x2$ . Then $dx=\frac{2}{1+t^2}dt,\quad \sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad 1+\cos x=\frac{2}{1+t^2}.$ Thus, $\sin x(1+\cos x)=\frac{2t}{1+t^2}\cdot \frac{2}{1+t^2}=\frac{4t}{(1+t^2)^2}.$ So,

=\frac{2+\frac{6t}{1+t^2}}{\frac{4t}{(1+t^2)^2}}\cdot \frac{2}{1+t^2}dt.$$ Simplify: $$=\frac{2(1+t^2)+6t}{1+t^2}\cdot \frac{(1+t^2)^2}{4t}\cdot \frac{2}{1+t^2}dt$$ $$=\frac{2(1+t^2+3t)}{1+t^2}\cdot \frac{1+t^2}{2t}dt$$ $$=\frac{1+t^2+3t}{t}dt$$ $$=\left(t+3+\frac{1}{t}\right)dt.$$ 6. Change the limits: - When $x=\pi/3$, $$t=\tan\frac{\pi}{6}=\frac{1}{\sqrt3}.$$ - When $x=\pi/2$, $$t=\tan\frac{\pi}{4}=1.$$ Therefore, $$I=\int_{1/\sqrt3}^{1}\left(t+3+\frac1t\right)dt.$$ 7. Integrate: $$I=\left[\frac{t^2}{2}+3t+\ln t\right]_{1/\sqrt3}^{1}.$$ Now evaluate: At $t=1$: $$\frac12+3+\ln 1=\frac72.$$ At $t=\frac{1}{\sqrt3}$: $$\frac{1}{2\cdot 3}+\frac{3}{\sqrt3}+\ln\left(\frac{1}{\sqrt3}\right) =\frac16+\sqrt3-\ln \sqrt3.$$ Hence, $$I=\frac72-\left(\frac16+\sqrt3-\ln\sqrt3\right)$$ $$=\frac{21-1}{6}-\sqrt3+\ln\sqrt3$$ $$=\frac{20}{6}-\sqrt3+\ln\sqrt3$$ $$=\frac{10}{3}-\sqrt3+\ln\sqrt3.$$ 8. Compare with options: $$\boxed{\frac{10}{3}-\sqrt3+\log_e\sqrt3}$$ This matches **Option A**.

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