Question

The integral ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}{\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx$ is equal to

• A. $\frac{9}{2}$
• B. $-\frac{1}{18}$
• C. $-\frac{1}{9}$
• D. $\frac{7}{18}$

Answer 1

The correct option is B

$-\frac{1}{18}$

Explanation for the correct answer:

Step 1: Reduce the given integral into a simpler form by using trigonometric identities

Given integral: ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}{\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx$.

Indefinite integral: $\int {\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx$

$$\begin{gathered} =\int {\tan }^{3}x·{\sin }^{2}3x\left[2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\left(2\sin 3x·\cos 3x\right)\right]dx\left[\because \sin 2x=2\sin x·\cos x\right] \\ =\int \left(2{\tan }^{3}x·{\sec }^{2}x·{\sin }^{4}3x+6{\tan }^{4}x·{\sin }^{3}3x·\cos 3x\right)dx \end{gathered}$$

Step 2: Solve the integral by substitution method

Let, $z={\tan }^{4}x·{\sin }^{4}3x$.

Differentiate both sides of the equation.

$$\begin{gathered} d\left(z\right)=d\left({\tan }^{4}x·{\sin }^{4}3x\right) \\ \Rightarrow dz={\sin }^{4}3x·d\left({\tan }^{4}x\right)+{\tan }^{4}x·d\left({\sin }^{4}3x\right) \\ \Rightarrow dz={\sin }^{4}3x·\left(4{\tan }^{3}x·{\sec }^{2}x\right)+{\tan }^{4}x·\left(4{\sin }^{3}3x·3\cos 3x\right) \\ \Rightarrow dz=4{\tan }^{3}x·{\sec }^{2}x·{\sin }^{4}3x+12{\tan }^{4}x·{\sin }^{3}3x·\cos 3x \\ \Rightarrow \frac{dz}{2}=2{\tan }^{3}x·{\sec }^{2}x·{\sin }^{4}3x+6{\tan }^{4}x·{\sin }^{3}3x·\cos 3x \end{gathered}$$

Thus, $\int {\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx=\int \left(2{\tan }^{3}x·{\sec }^{2}x·{\sin }^{4}3x+6{\tan }^{4}x·{\sin }^{3}3x·\cos 3x\right)dx$

$$\begin{gathered} =\int \frac{dz}{2} \\ =\frac{z}{2}+c\left[\because c=\mathrm{Constant}\mathrm{of}\mathrm{Integration}\right] \\ =\frac{{\tan }^{4}x·{\sin }^{4}3x}{2}+c \end{gathered}$$

Step 3: Apply the limits to integration to find the required value

Now, consider the upper and lower limits of the integral to solve the definite integral.

$$\begin{gathered} {\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}{\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx={\left[\frac{{\tan }^{4}x·{\sin }^{4}3x}{2}\right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ =\frac{1}{2}\left[{\tan }^4\left(\frac{\mathrm{\pi }}{3}\right)·{\sin }^4\left(3\times \frac{\mathrm{\pi }}{3}\right)-{\tan }^4\left(\frac{\mathrm{\pi }}{6}\right)·{\sin }^4\left(3\times \frac{\mathrm{\pi }}{6}\right)\right] \\ =\frac{1}{2}\left[{\left(\sqrt{3}\right)}^4·{\left(0\right)}^4-{\left(\frac{1}{\sqrt{3}}\right)}^4·{\left(1\right)}^4\right] \\ =\frac{1}{2}\left[0-\left(\frac{1}{9}\right)·1\right] \\ =-\frac{1}{18} \end{gathered}$$

Therefore, the integral ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}{\tan }^{3}x·{\sin }^{2}3x\left(2{\sec }^{2}x·{\sin }^{2}3x+3\tan x·\sin 6x\right)dx$ is equal to $-\frac{1}{18}$.