Solution : I = ∫_0^π /2 cosx dx #160; = [sinx]_0^π /2 #160; = [sinπ2 sin0] = 1 ...

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First Fundamental Theorem of Calculus

∫0π /2 cos x ...

Question

$π / 2 \int 0 cos x d x$

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Solution

Solution :-

$I = \int_{0}^{π / 2} c o s x d x$

$= [s i n x]_{0}^{π / 2}$

$= [s i n \frac{π}{2} - s i n 0]$

$= 1$

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Similar questions

\int_{0}^{\frac{π}{2}} \frac{cos x d x}{cos x + sin x}

I_{1} = \int_{0}^{π / 2} cos (sin x) d x

I_{2} = \int_{0}^{π / 2} sin (cos x) d x

I_{3} = \int_{0}^{π / 2} cos x d x

π / 2 \int 0 \frac{cos x d x}{(cos x + sin x)}

\int_{0}^{\frac{π}{2}} cos x d x = a

\int_{0}^{π} cos x d x =

\int_{0}^{π / 2} (cos x) d x = 1

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