What is the integral from 0 to pi/2 of e^{sin(5pix)}cos(5pix) ?

The integral from 0 to pi/2 of e^{sin(5pix)}cos(5pix) is -(1-e^{sin((5pi^2)/2)})/(5pi)

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integral from 0 to pi/2 of e^{sin(5pix)}cos(5pix)

\int_{0}^{\frac{π}{2}} e^{s i n (5 π x)} cos (5 π x) d x

- \frac{1 - e ^{s i n (\frac{5 π ^{2}}{2})}}{5 π}

Decimal

- 0.02277 \dots

Solution steps

\int_{0}^{\frac{π}{2}} e^{s i n (5 π x)} cos (5 π x) d x

Apply u-substitution

: \int_{1}^{e^{s i n (\frac{5 π ^{2}}{2})}} \frac{1}{5 π} d u

= \int_{1}^{e^{s i n (\frac{5 π ^{2}}{2})}} \frac{1}{5 π} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{e^{s i n (\frac{5 π ^{2}}{2})}}^{1} \frac{1}{5 π} d u

Integral of a constant:

\int a d x = a x

= - [\frac{1}{5 π} u]_{e^{s i n (\frac{5 π ^{2}}{2})}}^{1}

Compute the boundaries:

\frac{1 - e ^{s i n (\frac{5 π ^{2}}{2})}}{5 π}

= - \frac{1 - e ^{s i n (\frac{5 π ^{2}}{2})}}{5 π}

\int 2 x (x^{2} + 6)^{7} d x

{\frac{1}{4 s + 1}}

\int (5 sec (x) - cos (x))^{2} d x

\int cot^{3} (4 x) sec^{2} (4 x) d x

\int (4 - x^{2}) d x

What is the integral from 0 to pi/2 of e^{sin(5pix)}cos(5pix) ?
The integral from 0 to pi/2 of e^{sin(5pix)}cos(5pix) is -(1-e^{sin((5pi^2)/2)})/(5pi)