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Question: How do you integrate \( \int {{{\sin }^{10}}x\cos xdx} \) using substitution?...

How do you integrate ∫sin⁡10xcos⁡xdx\int {{{\sin }^{10}}x\cos xdx} using substitution?

Explanation

Solution

In this question we have to evaluate the given integral, by introducing a new independent variable when it is difficult to find the integration of the function. By changing the independent variable x to m, in a given form of integral function say (∫f(x))\left( {\int {f(x)} } \right) , we can transform the function and then find it’s integral value.

Complete step by step solution:
Given Integral
I=∫sin⁡10xcos⁡xdxI = \int {{{\sin }^{10}}x\cos xdx}
As we do not have any direct integral formula for higher order functions of trigonometric ratios, therefore, we use substitution methods.
So Put sin⁡x=t\sin x = t
Differentiate both sides of the equation with respect to x
d(sin⁡x)dx=dtdx\dfrac{{d(\sin x)}}{{dx}} = \dfrac{{dt}}{{dx}}
⇒−cos⁡x=dtdx\Rightarrow - \cos x = \dfrac{{dt}}{{dx}}
⇒cos⁡xdx=−dt\Rightarrow \cos xdx = - dt
Substitute the above derived result in the given integral, we have
I=−∫t10dtI = - \int {{t^{10}}dt}
As we know that ∫xndx=xn+1n\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{n}} so, using this property we get
I=−t10+110I = - \dfrac{{{t^{10 + 1}}}}{{10}}
I=−t1110I = - \dfrac{{{t^{11}}}}{{10}}
Now substitute the value of t back in the integral to get the answer in terms of x
⇒I=−sin⁡11x11+C\Rightarrow I = - \dfrac{{{{\sin }^{11}}x}}{{11}} + C
Thus, this is the required answer.
So, the correct answer is “ I=−sin⁡11x11+C I = - \dfrac{{{{\sin }^{11}}x}}{{11}} + C ”.

Note : Whenever we are required to solve these type of questions the key concept is to simplify the inside entities of the integration to the basic level either by performing elementary algebraic operations or by substitution, whichever suits the conditions of the question, so that the direct integration formulas of the so obtained result can be applied.