#### Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 38 textbook solution.

Hints:-  You must know the integral rules of trignometric functions and its limits.

Given:-  $\int_{0}^{2 \pi} \sin ^{100} x \cos ^{|10|} x \cdot d x$

Solution : $\int_{0}^{2 \pi} \sin ^{100} x \cos ^{|10|} x \cdot d x$

Suppose :  $f(x)=\sin ^{100} x \cos ^{\left |1 0 \right |} x$

Now $f(2 \pi-x)=\sin ^{100}(2 \pi-x) \cos ^{\left |1 0 \right |}(2 \pi-x)$

\begin{aligned} &=(-\sin x)^{100}(\cos x)^{|10|} \\ &=\sin ^{100} x \cdot \cos ^{|10|} x \\ &=f(x) \end{aligned}

\begin{aligned} \therefore I &=\int_{0}^{2 \pi} \sin ^{100} x \cos ^{10} x \cdot d x \\ &=2 \int_{0}^{\pi} \sin ^{100} x \cos ^ {10} x \cdot d x \end{aligned}

Again

\begin{aligned} f(\pi-x) &=\sin ^{100}(\pi-x) \cdot \cos ^{|10|}(\pi-x) \\ &=\sin ^{100} x \cdot(-\cos x)^{|10|}x \\ &=-\sin ^{100} x \cdot \cos ^{10}x \\ &=-f(x) \end{aligned}

Its and odd function

\begin{aligned} \therefore I &=2 \times 0 \\ &=0 \end{aligned}