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  • Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.28 Question 5 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.28 Question 5 Maths Textbook Solution.

Answers (1)

Answer:-

\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1} \frac{\sin x}{2}+c

Hint:-

Taking \sin x = t

Given:-

\int \cos x \sqrt{4-\sin ^{2} x} d x

Solution:-

By taking \sin x = t

\cos x d x=d t

Hence

\begin{aligned} &=\int \sqrt{4-t^{2}} d t \\\\ &=\int \sqrt{(2)^{2}-(t)^{2}} d t \\\\ &=\frac{1}{2} t \sqrt{4-t^{2}}+\frac{1}{2}(2)^{2} \sin ^{-1} \frac{t}{2}+c \end{aligned}

 

Using the formula

\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &=\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1} \frac{\sin x}{2}+c \end{aligned}

 

 

 

 

Posted by

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