int x^{2} [ sqrt{2} sin ( frac{pi}{4} + x ) + | vedclass

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(A) समाकल व्यंजक इस प्रकार है,$I = \int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \sin \frac{

Vedclass · Accepted Answer

$( x^{2} + 2 x - 2 ) \sin x + ( - x^{2} + 2 x + 2 ) \cos x + ( x^{2} - 2 x + 2 ) e^{x} + C$

Vedclass · Answer

(A) समाकल व्यंजक इस प्रकार है,$I = \int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \sin \frac{\pi}{4} \cos x + \sin x \cos \frac{\pi}{4} ) + e^{x} ] dx$ $\Rightarrow I = \int x^{2} [ \sqrt{2} ( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ) + e^{x} ] dx$ सरल करने पर,$\Rightarrow I = \int x^{2} ( \cos x + \sin x + e^{x} ) dx$ $\Rightarrow I = \int x^{2} ( \cos x + \sin x ) dx + \int x^{2} e^{x} dx$ खंडशः समाकलन का उपयोग करते हुए,$\int u v dx = u \int v dx - \int \{ \frac{d u}{d x} \cdot \int v dx \} dx$: $\Rightarrow I = x^{2} ( \sin x - \cos x ) - \int 2 x ( \sin x - \cos x ) dx + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = x^{2} ( \sin x - \cos x ) - 2 [ x ( - \cos x - \sin x ) - \int ( - \cos x - \sin x ) dx ] + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = x^{2} ( \sin x - \cos x ) + 2 x ( \cos x + \sin x ) - 2 ( \sin x - \cos x ) + ( x^{2} - 2 x + 2 ) e^{x} + C$ $\Rightarrow I = ( x^{2} + 2 x - 2 ) \sin x + ( - x^{2} + 2 x + 2 ) \cos x + ( x^{2} - 2 x + 2 ) e^{x} + C$

$\int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx =$

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