What is the integral from 1 to cos(x) of (t+sin(t)) ?

The integral from 1 to cos(x) of (t+sin(t)) is cos(1)-cos(cos(x))-(sin^2(x))/2

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integral from 1 to cos(x) of (t+sin(t))

\int_{1}^{c o s (x)} (t + sin (t)) d t

cos (1) - cos (cos (x)) - \frac{sin ^{2} ( x )}{2}

Solution steps

\int_{1}^{c o s (x)} t + sin (t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{c o s (x)} t d t + \int_{1}^{c o s (x)} sin (t) d t

\int_{1}^{c o s (x)} t d t = \frac{cos ^{2} ( x ) - 1}{2}

\int_{1}^{c o s (x)} sin (t) d t = - cos (cos (x)) + cos (1)

= \frac{cos ^{2} ( x ) - 1}{2} - cos (cos (x)) + cos (1)

Simplify

\frac{cos ^{2} ( x ) - 1}{2} - cos (cos (x)) + cos (1) : cos (1) - cos (cos (x)) - \frac{sin ^{2} ( x )}{2}

= cos (1) - cos (cos (x)) - \frac{sin ^{2} ( x )}{2}

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What is the integral from 1 to cos(x) of (t+sin(t)) ?
The integral from 1 to cos(x) of (t+sin(t)) is cos(1)-cos(cos(x))-(sin^2(x))/2