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Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 49 Maths Textbook Solution.

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\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx


\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx=\int_{0}^{1} xe^{x}dx+\int_{0}^{1}\cos \frac{\pi x}{4}dx

Applying integration by parts method in !st integral then

=\left[x \int e^{x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{x} d x\right) d x+\left[\frac{\sin \frac{\pi}{4} x}{\frac{\pi}{4}}\right]_{0}^{1}

=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x+\frac{4}{\pi}\left[\sin \frac{\pi x}{4}\right]_{0}^{1}

=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{x}\right]_{0}^{1}+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

=[e-0]-\left[e^{1}-e^{0}\right]+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

= \left [ e \right ]-[e-1]+\frac{4}{\pi}\left[\frac{1}{\sqrt{2}}-0\right]

= e-e+1+\frac{4}{\pi} \times \frac{1}{\sqrt{2}}

=\frac{2\sqrt{2}}{\pi }+1

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