CUET UG 2025 — Mathematics Arithmetic

When integrating $e^{ax}$eax multiplied with a combination of $\sin x$sinx and $\cos x$cosx, the result typically has the form $e^{ax}$eax times another trigonometric combination.

For this integral, consider the function $\frac{1}{2}e^{2x}\sin x$21​e2xsinx.

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To check if this is correct, differentiate $\frac{1}{2}e^{2x}\sin x$21​e2xsinx using the product rule:

$$\frac{d}{dx}\left[\frac{1}{2}e^{2x}\sin x\right]$$dxd​[21​e2xsinx]

$$= \frac{1}{2}\left[(e^{2x})' \cdot \sin x + e^{2x} \cdot (\sin x)'\right]$$=21​[(e2x)′⋅sinx+e2x⋅(sinx)′]

$$= \frac{1}{2}\left[2e^{2x} \cdot \sin x + e^{2x} \cdot \cos x\right]$$=21​[2e2x⋅sinx+e2x⋅cosx]

$$= e^{2x}\sin x + \frac{1}{2}e^{2x}\cos x$$=e2xsinx+21​e2xcosx

$$= e^{2x}\left(\sin x + \frac{1}{2}\cos x\right)$$=e2x(sinx+21​cosx)

This matches the integrand exactly.

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Therefore:

$$\int e^{2x}\left(\sin x + \frac{1}{2}\cos x\right) dx = \frac{1}{2}e^{2x}\sin x + C$$∫e2x(sinx+21​cosx)dx=21​e2xsinx+C