Question

What is the value of the integral $$\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}{\sin }^{-4}xdx$$?
A. $${\displaystyle \frac{3}{2}}$$
B. $$-{\displaystyle \frac{8}{3}}$$
C. $${\displaystyle \frac{3}{8}}$$
D. $${\displaystyle \frac{8}{3}}$$

Answer 1

Hint: Here, a definite integral is given. First, rewrite the given term $${\sin }^{-4}x$$ as $${\displaystyle \frac{1}{{\sin }^{4}x}}$$. Then, multiply the numerator and the denominator by $${\cos }^{4}x$$ and simplify the integral. Then, simplify the integral by using the trigonometric ratios. After that, apply the integration rule for the limit of the integral $$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$$. Then simplify the numerator by using the trigonometric formula $$1+{\tan }^{2}x={\sec }^{2}x$$. Now, substitute $$\tan x=u$$ in the given integral and solve it by using the integration formulas. In the end, apply the upper and lower limit of the integration and solve it to get the required answer.


Formula Used: $$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$$, if $$f\left(x\right)$$ is an even function
$${\displaystyle \frac{1}{\cos x}}=\sec x$$
$${\displaystyle \frac{\sin x}{\cos x}}=\tan x={\displaystyle \frac{1}{\cot x}}$$
$$1+{\tan }^{2}x={\sec }^{2}x$$
$$\int x^{-n}dx={\displaystyle \frac{x^{-n+1}}{-n+1}}$$

Complete step by step solution: The given integral is $$\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}{\sin }^{-4}xdx$$.
Let consider,
$$I=\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}{\sin }^{-4}xdx$$
Let’s simplify the above integral.
Rewrite the given term $${\sin }^{-4}x$$ as $${\displaystyle \frac{1}{{\sin }^{4}x}}$$.
$$I=\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}{\displaystyle \frac{1}{{\sin }^{4}x}}dx$$
Now multiply the numerator and the denominator of the right-hand side by $${\cos }^{4}x$$.
$$I=\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left[{\displaystyle \frac{{\cos }^{4}x}{{\sin }^{4}x}}\times {\displaystyle \frac{1}{{\cos }^{4}x}}\right]dx$$
Simplify the above integral by using the basic trigonometric ratios.
$$I=\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left[{\displaystyle \frac{1}{{\tan }^{4}x}}{\sec }^{4}x\right]dx$$
Now apply the integration rule for the limit.
$$I=2\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left[{\displaystyle \frac{{\sec }^{4}x}{{\tan }^{4}x}}\right]dx$$
$$\Rightarrow I=2\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left[{\displaystyle \frac{\left({\sec }^{2}x\right)\left({\sec }^{2}x\right)}{{\tan }^{4}x}}\right]dx$$
Apply the trigonometric formula $$1+{\tan }^{2}x={\sec }^{2}x$$
$$\Rightarrow I=2\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left[{\displaystyle \frac{\left(1+{\tan }^{2}x\right)\left({\sec }^{2}x\right)}{{\tan }^{4}x}}\right]dx$$ $$.....\left(1\right)$$
Now substitute $$\tan x=u$$ in the above equation.
Differentiate the substituting equation, we get
$${\sec }^{2}xdx=du$$
And limits changes as follows:
$$x=0\Rightarrow u=0$$ and $$x={\displaystyle \frac{\pi }{4}}\Rightarrow u=1$$

We get the equation $$\left(1\right)$$ as follows:
$$\Rightarrow I=2\underset{0}{\overset{1}{\int }}{\displaystyle \frac{1+u^2}{u^4}}du$$
Simplify the right-hand side.
$$\Rightarrow I=2\underset{0}{\overset{1}{\int }}\left[{\displaystyle \frac{1}{u^4}}+{\displaystyle \frac{1}{u^2}}\right]du$$
$$\Rightarrow I=2\underset{0}{\overset{1}{\int }}\left[u^{-4}+u^{-2}\right]du$$
Apply the addition rule of integration.
$$\Rightarrow I=2\left[\underset{0}{\overset{1}{\int }}{u}^{-4}du+\underset{0}{\overset{1}{\int }}{u}^{-2}du\right]$$
Solve both integrals by using the rule $$\int x^{-n}dx={\displaystyle \frac{x^{-n+1}}{-n+1}}$$.
$$\Rightarrow I=2{\left[{\displaystyle \frac{u^{-4+1}}{-4+1}}+{\displaystyle \frac{u^{-2+1}}{-2+1}}\right]}_0^1$$
$$\Rightarrow I=2{\left[{\displaystyle \frac{u^{-3}}{-3}}+{\displaystyle \frac{u^{-1}}{-1}}\right]}_0^1$$
$$\Rightarrow I=-2\left[{\left|{\displaystyle \frac{1}{3u^3}}\right|}_0^1+{\left|{\displaystyle \frac{1}{u}}\right|}_0^1\right]$$
$$\Rightarrow I=-2\left[\left|{\displaystyle \frac{1}{3{\left(1\right)}^3}}-{\displaystyle \frac{1}{3{\left(0\right)}^3}}\right|+\left|{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{0}}\right|\right]$$
$$\Rightarrow I=-2\left[{\displaystyle \frac{1}{3}}+1\right]$$
$$\Rightarrow I=-2\left[{\displaystyle \frac{4}{3}}\right]$$
$$\Rightarrow I=-{\displaystyle \frac{8}{3}}$$
Therefore,
$$\underset{-{\displaystyle \frac{\pi }{4}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}{\sin }^{-4}xdx=-{\displaystyle \frac{8}{3}}$$

Option ‘B’ is correct

Note: Students often get confused about the formula of the definite integral of the function. They used $$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=F\left(b\right)+F\left(a\right)$$ , which is incorrect. The correct formula is $$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$$.
Sometimes they also add integration constant $$c$$ in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.