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  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 45 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 45 maths

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Answer: \frac{\sqrt{2}-1}{\sqrt{2}}

Hint: you must know the rule of integration

Given: \int_{0}^{\frac{\pi}{4}} \sin \{x\} d x

Solution:  \int_{0}^{\frac{\pi}{4}} \sin \{x\} d x

\begin{aligned} &=-[\cos \{x\}]_{0}^{\frac{\pi}{4}} \\\\ &=-\left[\cos \frac{\pi}{4}-\cos 0\right] \end{aligned}

\begin{aligned} &=-\left[\frac{1}{\sqrt{2}}-1\right] \\\\ &=\frac{\sqrt{2}-1}{\sqrt{2}} \end{aligned}

 

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