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Question

Find the value of the definite integral $$I=\displaystyle \int_0^{\pi}|\sqrt{2}\sin x+2\cos x|dx$$.


Solution

Consider the given integral.

 $$I=\int_{0}^{\pi }{|\sqrt{2}\sin x+2\cos x|}dx$$

 

Solve the definite integration.

 

 $$ I=\int_{0}^{\pi }{|\sqrt{2}\sin x+2\cos x|}dx $$

 $$ =\int_{0}^{\frac{\pi }{2}}{(\sqrt{2}\sin x+2\cos x)}dx+\int_{\frac{\pi }{2}}^{\pi }{(-\sqrt{2}\sin x-2\cos x)}dx $$

 $$ =\int_{0}^{\frac{\pi }{2}}{(\sqrt{2}\sin x+2\cos x)}dx+\int_{\frac{\pi }{2}}^{\pi }{(\sqrt{2}\sin x+2\cos x)}dx $$

 $$ =\left( -\sqrt{2}\cos x+2\sin x \right)_{0}^{\frac{\pi }{2}}+\left( -\sqrt{2}\cos x+2\sin x \right)_{\frac{\pi }{2}}^{\pi } $$

$$=\left( \left( -\sqrt{2}\cos \dfrac{\pi }{2}+2\sin \dfrac{\pi }{2} \right)-\left( -\sqrt{2}\cos 0+2\sin 0 \right) \right)+\left( \left( -\sqrt{2}\cos \dfrac{\pi }{2}+2\sin \dfrac{\pi }{2} \right)-\left( -\sqrt{2}\cos \pi +2\sin \pi  \right) \right) $$

 $$ =6.82 $$

Hence, this is the required result.


Mathematics

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