If \theta 1 and \theta 2 be respectively the smallest and the largest values of \theta in (0, 2 \pi ) - \pi which satisfy the equation, 2cot2 \theta - 5 \over \sin \theta + 4 = 0, then \int\limits_\theta _1^\theta _2 \cos ^23\theta d\theta is equal to :

Definite Integration PYQ — JEE Main 2020 · 7 Jan · Shift 2 · Q33 with Solution

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Correct answer: C

Solve the trigonometric equation

We need the values of $\theta \in (0,2\pi)\setminus\{\pi\}$ satisfying $2\cot^2\theta-\frac{5}{\sin\theta}+4=0.$

Using $\cot^2\theta=\frac{\cos^2\theta}{\sin^2\theta}=\frac{1-\sin^2\theta}{\sin^2\theta},$ let $s=\sin\theta \quad (s\neq 0).$

Then $2\frac{1-s^2}{s^2}-\frac{5}{s}+4=0.$ Multiply by $s^2$ : $2(1-s^2)-5s+4s^2=0.$ So, $2-2s^2-5s+4s^2=0$ $2s^2-5s+2=0.$ Factorizing: $(2s-1)(s-2)=0.$

Since $\sin\theta=s\in[-1,1]$ , we must have $s=\frac12.$ Thus, $\sin\theta=\frac12.$

In $(0,2\pi)\setminus\{\pi\}$ , the solutions are $\theta=\frac{\pi}{6},\ \frac{5\pi}{6}.$ Hence, $\theta_1=\frac{\pi}{6},\qquad \theta_2=\frac{5\pi}{6}.$

Evaluate the integral

We need

=\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta.$$ Use $$\cos^2 x=\frac{1+\cos 2x}{2}.$$ So, $$I=\int_{\pi/6}^{5\pi/6}\frac{1+\cos 6\theta}{2}\,d\theta =\frac12\int_{\pi/6}^{5\pi/6}1\,d\theta+\frac12\int_{\pi/6}^{5\pi/6}\cos 6\theta\,d\theta.$$ Now, $$\frac12\int_{\pi/6}^{5\pi/6}1\,d\theta =\frac12\left(\frac{5\pi}{6}-\frac{\pi}{6}\right) =\frac12\cdot\frac{4\pi}{6} =\frac{\pi}{3}.$$ And $$\frac12\int_{\pi/6}^{5\pi/6}\cos 6\theta\,d\theta =\frac12\left[\frac{\sin 6\theta}{6}\right]_{\pi/6}^{5\pi/6} =\frac{1}{12}\left(\sin 5\pi-\sin \pi\right)=0.$$ Therefore, $$I=\frac{\pi}{3}.$$ --- 3. **Check options** The value is $$\boxed{\frac{\pi}{3}}$$ which corresponds to **Option C**. --- 4. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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View written solutionFree

Correct answer: C

Solve the trigonometric equation

We need the values of $\theta \in (0,2\pi)\setminus\{\pi\}$ satisfying $2\cot^2\theta-\frac{5}{\sin\theta}+4=0.$

Using $\cot^2\theta=\frac{\cos^2\theta}{\sin^2\theta}=\frac{1-\sin^2\theta}{\sin^2\theta},$ let $s=\sin\theta \quad (s\neq 0).$

Then $2\frac{1-s^2}{s^2}-\frac{5}{s}+4=0.$ Multiply by $s^2$ : $2(1-s^2)-5s+4s^2=0.$ So, $2-2s^2-5s+4s^2=0$ $2s^2-5s+2=0.$ Factorizing: $(2s-1)(s-2)=0.$

Since $\sin\theta=s\in[-1,1]$ , we must have $s=\frac12.$ Thus, $\sin\theta=\frac12.$

In $(0,2\pi)\setminus\{\pi\}$ , the solutions are $\theta=\frac{\pi}{6},\ \frac{5\pi}{6}.$ Hence, $\theta_1=\frac{\pi}{6},\qquad \theta_2=\frac{5\pi}{6}.$

Evaluate the integral

We need

=\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta.$$ Use $$\cos^2 x=\frac{1+\cos 2x}{2}.$$ So, $$I=\int_{\pi/6}^{5\pi/6}\frac{1+\cos 6\theta}{2}\,d\theta =\frac12\int_{\pi/6}^{5\pi/6}1\,d\theta+\frac12\int_{\pi/6}^{5\pi/6}\cos 6\theta\,d\theta.$$ Now, $$\frac12\int_{\pi/6}^{5\pi/6}1\,d\theta =\frac12\left(\frac{5\pi}{6}-\frac{\pi}{6}\right) =\frac12\cdot\frac{4\pi}{6} =\frac{\pi}{3}.$$ And $$\frac12\int_{\pi/6}^{5\pi/6}\cos 6\theta\,d\theta =\frac12\left[\frac{\sin 6\theta}{6}\right]_{\pi/6}^{5\pi/6} =\frac{1}{12}\left(\sin 5\pi-\sin \pi\right)=0.$$ Therefore, $$I=\frac{\pi}{3}.$$ --- 3. **Check options** The value is $$\boxed{\frac{\pi}{3}}$$ which corresponds to **Option C**. --- 4. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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