Question

Evaluate the following and then find the value: $\int{{{\cos }^{-3/7}}x{{\sin }^{-11/7}}xdx}$ is equal to
$\left( A \right)\text{ }\log \left| {{\sin }^{4/7}}x \right|+c$
$\left( B \right)\text{ }\dfrac{4}{7}{{\tan }^{4/7}}x+c$
$\left( C \right)\text{ }\dfrac{-7}{4}{{\tan }^{-4/7}}x+c$
$\left( D \right)\text{ }\log {{\cos }^{3/7}}x+c$

Answer 1

In this question we have been given with the integration of a trigonometric expression. We will use the law of exponents in this sum such as the formulas ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}$, ${{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}$ and simplify the expression by using these results. We will then use the method of substitution and change the expression and integrate. After integration we will re-substitute the value to get the required solution.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \int{{{\cos }^{-3/7}}x{{\sin }^{-11/7}}xdx}$
Now we know that $\dfrac{-11}{7}$ can be written as $-2+\dfrac{3}{7}$ therefore, on substituting in the expression, we get:
$\Rightarrow \int{{{\cos }^{\dfrac{-3}{7}}}x{{\sin }^{-2+\dfrac{3}{7}}}xdx}$
Now we can use the property of exponents that ${{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}$ therefore, we can write the expression as:
$\Rightarrow \int{{{\cos }^{\dfrac{-3}{7}}}x{{\sin }^{-2}}x{{\sin }^{\dfrac{3}{7}}}xdx}$
Now we know that ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}$ therefore, we can write the expression as:
$\Rightarrow \int{\dfrac{{{\cos }^{\dfrac{-3}{7}}}x{{\sin }^{\dfrac{3}{7}}}x}{{{\sin }^{2}}x}dx}$
Now we know that $\dfrac{1}{{{\sin }^{2}}x}={{\csc }^{2}}x$ therefore, on substituting it in the expression, we get:
$\Rightarrow \int{{{\cos }^{\dfrac{-3}{7}}}x{{\sin }^{\dfrac{3}{7}}}x{{\csc }^{2}}xdx}$
Now on using ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}$ therefore, we can write the expression as:
$\Rightarrow \int{\dfrac{{{\sin }^{\dfrac{3}{7}}}x{{\csc }^{2}}x}{{{\cos }^{\dfrac{3}{7}}}x}dx}$
Now we know that $\dfrac{\sin x}{\cos x}=\tan x$ therefore, we get:
$\Rightarrow \int{{{\tan }^{\dfrac{3}{7}}}x{{\csc }^{2}}xdx}$
Now we know that $\dfrac{1}{\cot x}=\tan x$ therefore, on substituting, we get:
$\Rightarrow \int{\dfrac{{{\csc }^{2}}x}{{{\cot }^{\dfrac{3}{7}}}x}dx}$
Now consider $\cot x=t$
On differentiating with respect to $x$, we get $-\csc xdx=dt$.
On substituting, we get:
$\Rightarrow -\int{\dfrac{dt}{{{t}^{\dfrac{3}{7}}}}}$
Now we can rewrite the expression as:
$\Rightarrow -\int{{{t}^{-\dfrac{3}{7}}}dt}$
Now we know the formula that $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ therefore on using the formula, we get:
$\Rightarrow -\int{\dfrac{{{t}^{-\dfrac{3}{7}+1}}}{-\dfrac{3}{7}+1}dt}$
On simplifying, we get:
$\Rightarrow -\int{\dfrac{{{t}^{\dfrac{4}{7}}}}{\dfrac{4}{7}}dt}$
On rearranging the expression, we get:
$\Rightarrow -\dfrac{7}{4}{{\operatorname{t}}^{4/7}}+C$
On substituting the value of $t$, we get:
$\Rightarrow -\dfrac{7}{4}{{\cot }^{4/7}}x+C$
Since $\tan x$ is the inverse of $\cot x$, we can write the expression as:
$\Rightarrow -\dfrac{7}{4}{{\tan }^{-4/7}}x+C$, which is the required solution therefore the correct answer is option $(C)$.

Note: It is to be remembered that in this question we have used the method of substitution to solve the problem. There also exist other methods such as partial fractions and integration by parts to solve the questions of integration hence, they should be remembered. In this question we have an indefinite integral which has no limit.