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Question: What is the value of \(\int\limits_0^\pi {{{\sin }^{50}}x{{\cos }^{49}}x} dx\) ?...

What is the value of ∫0πsin⁡50xcos⁡49xdx\int\limits_0^\pi {{{\sin }^{50}}x{{\cos }^{49}}x} dx ?

Explanation

Solution

Start by taking the given integral as II . Look for the upper limit and lower limit and then try to use the properties of definite integral which suits the best and can make the problem easier, find out the value of the given integral using this property.

Complete step-by-step solution:
Let I=∫0πsin⁡50xcos⁡49xdxI = \int\limits_0^\pi {{{\sin }^{50}}x{{\cos }^{49}}x} dx ...…(1)
We know the property
∫02af(x)dx= i)2∫0af(x)dx,if f(2a−x)=f(x) ii)0,if f(2a−x)=−f(x)  \int\limits_0^{2a} {f(x)} dx = \\\ i)2\int\limits_0^a {f(x)} dx,{\text{if }}f(2a - x) = f(x) \\\ ii)0,{\text{if }}f(2a - x) = - f(x) \\\
So , let us find f(2a−x)f(2a - x)
f(x)=sin⁡50xcos⁡49x f(2a−x)=f(π−x) ⇒f(π−x)=sin⁡50(π−x)cos⁡49(π−x) ⇒f(π−x)=−sin⁡50(x)cos⁡49(x) ⇒f(π−x)=−f(x)  f(x) = {\sin ^{50}}x{\cos ^{49}}x \\\ f(2a - x) = f(\pi - x) \\\ \Rightarrow f(\pi - x) = {\sin ^{50}}(\pi - x){\cos ^{49}}(\pi - x) \\\ \Rightarrow f(\pi - x) = - {\sin ^{50}}(x){\cos ^{49}}(x) \\\ \Rightarrow f(\pi - x) = - f(x) \\\
So we have got condition (ii),
∴I=0\therefore I = 0

Note: This type of question is generally done by using the property of definite integration. There are many other properties of definite integral, must be practiced by the students very well, as choosing the right property would help in making the problem very easy and quick to solve. Also, Attention must be given to trigonometric functions and their periodicity, if any, while changing limits or function.