[3] ( int_{0}^{frac{3 pi}{2}}(5 cos x-2 sin x) d x )

Question

Accepted Answer

Step:1 ：( int_{0}^{frac{3 pi}{2}}(5 cos x-2 sin x) dx = 5 int_{0}^{frac{3 pi}{2}} cos x dx - 2 int_{0}^{frac{3 pi}{2}} sin x dx )Step:2 ：(= 5 [sin x]_{0}^{frac{3 pi}{2}} - 2 [-cos x]_{0}^{frac{3 pi}{2}} )Step:3 ：(= 5( sin (frac{3 pi}{2}) - sin 0) - 2 (- cos (frac{3 pi}{2}) + cos 0) )Step:4 ：(= 5(-1 - 0) - 2 (0 + 1) )Step:5 ：(= -5 + 2 )

Answer

Step: 1 ：( int_{0}^{frac{3 pi}{2}}(5 cos x-2 sin x) dx = 5 int_{0}^{frac{3 pi}{2}} cos x dx - 2 int_{0}^{frac{3 pi}{2}} sin x dx )Step: 2 ：(= 5 [sin x]_{0}^{frac{3 pi}{2}} - 2 [-cos x]_{0}^{frac{3 pi}{2}} )Step: 3 ：(= 5( sin (frac{3 pi}{2}) - sin 0) - 2 (- cos (frac{3 pi}{2}) + cos 0) )Step: 4 ：(= 5(-1 - 0) - 2 (0 + 1) )Step: 5 ：(= -5 + 2 )

[3] \( \int_{0}^{\frac{3 \pi}{2}}(5 \cos x-2 \sin x) d x \)

Answer

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