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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.6
Integrate the functions in Exercises 134 to 155.
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Question : 154 of 271
Marks: +1, -0
e2xsin⁡xe^{2x} \sin x
Solution:  
Let I=∫e2xsin⁡x dxI=\int e^{2x} \sin x \, dx
⇒I=e2x∫sin⁡x dx−∫(ddxe2x⋅∫sin⁡x dx)dx\Rightarrow I=e^{2x}\int \sin x \, dx - \int \left( \frac{d}{dx} e^{2x} \cdot \int \sin x \, dx \right) dx
⇒I=e2x(−cos⁡x)−∫2e2x(−cos⁡x) dx+C1\Rightarrow I=e^{2x}(-\cos x)-\int 2e^{2x}(-\cos x)\,dx+C_1
⇒I=−e2xcos⁡x+2∫e2xcos⁡x dx+C1\Rightarrow I=-e^{2x}\cos x+2\int e^{2x}\cos x\,dx+C_1
⇒I=−e2xcos⁡x+2[e2x∫cos⁡x dx−∫(ddxe2x⋅∫cos⁡x dx)dx]+C1\Rightarrow I=-e^{2x}\cos x+2\left[ e^{2x}\int \cos x\,dx - \int \left( \frac{d}{dx} e^{2x} \cdot \int \cos x\,dx \right) dx \right]+C_1
⇒I=−e2xcos⁡x+2e2xsin⁡x−4∫e2xsin⁡x dx+C1+C2\Rightarrow I=-e^{2x}\cos x+2e^{2x}\sin x-4\int e^{2x}\sin x\,dx+C_1+C_2
⇒I=e2x(2sin⁡x−cos⁡x)−4I+C1+C2\Rightarrow I= e^{2x} (2\sin x-\cos x)-4I+C_1+C_2
⇒5I=e2x(2sin⁡x−cos⁡x)+C1+C2\Rightarrow 5I= e^{2x} (2\sin x-\cos x)+C_1+C_2
⇒I=e2x5(2sin⁡x−cos⁡x)+C\Rightarrow I= \frac{e^{2x}}{5} (2\sin x-\cos x)+C, where C+C1+C2C+C_1+C_2
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