To evaluate this integral, we must first rewrite the $\displaystyle \sec$ from the denominator, as a $\displaystyle \cos$ in the numerator. This gives us

$\displaystyle \int_0^{\pi/2} \sin^2(x)\cos(x)dx$

Since the $\displaystyle \sin$ has an exponent, we will pick it for our u-substitution.

$\displaystyle u = \sin(x)$

Now we find $\displaystyle du$ by differentiating.

$\displaystyle du = \cos(x)dx$

This gives us the cosine piece. Writing everything in terms or u, we get

$\displaystyle \int_{u_a}^{u_b} u^2du$

To integrate this we use the following basic integral form:$\displaystyle \int_a^b u^n du = [\frac{1}{n+1}u^{n+1}]|_a^b$

Applying this to our integral, we get

$\displaystyle [\frac{1}{3}u^3]$

Now we can "un-substitute" to get back to x-terms. Replacing $\displaystyle u$ with $\displaystyle \sin(x)$, we get

$\displaystyle [\frac{1}{3}\sin^3(x)]|_0^{\pi/2}$

Now we use the 2nd Fundamental Theorem of Calculus. We plug in the upper bound of $\displaystyle \pi/2$ , then plug in the lower bound of $\displaystyle 0$, and subtract.

Doing this we get

$\displaystyle [\frac{1}{3}\sin^3(\pi/2)]-[\frac{1}{3}\sin^3(0)]$

$\displaystyle [\frac{1}{3}(1^3)]-[\frac{1}{3}(0^3)]$

$\displaystyle \frac{1}{3}(1)-\frac{1}{3}(0)$

$\displaystyle \frac{1}{3}$

This is the correct answer.

We proceed as follows

$\displaystyle \int_0^{10} f(x) dx$. (Start)

$\displaystyle =\int_0^4 f(x) dx + \int_4^{10}f(x)dx$. (Break up the integral using the additive rule.)

$\displaystyle =\int_0^4 f(x) dx + \int_4^{10}2g(x)dx$. (We don't have information about the 2nd integral, so we solve our first equation for $\displaystyle f(x)$ and replace it in this integral.)

$\displaystyle =\int_0^4 f(x) dx + 2\int_4^{10}g(x)dx$. (Factor out the $\displaystyle 2$ using linearity.)

$\displaystyle =6 +2(8)$. (Substitute in what we were given.)

$\displaystyle =22$.

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{9}^{34}f(x)dx\\&\int_{9}^{34}f(x)dx=-(-24)+(11)+(24)+(34)\\&\int_{9}^{34}f(x)dx=93\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{2}^{31}f(x)dx\\&\int_{2}^{31}f(x)dx=-\int_{13}^{2}f(x)dx+\int_{13}^{19}f(x)dx-\int_{23}^{19}f(x)dx-\int_{28}^{23}f(x)dx-\int_{31}^{28}f(x)dx\\&\int_{2}^{31}f(x)dx=-(-94)+(-23)-(-32)-(-18)-(-84)\\&\int_{2}^{31}f(x)dx=205\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-10}^{28}f(x)dx\\&\int_{-10}^{28}f(x)dx=\int_{-10}^{-5}f(x)dx-\int_{-1}^{-5}f(x)dx-\int_{6}^{-1}f(x)dx-\int_{21}^{6}f(x)dx-\int_{28}^{21}f(x)dx\\&\int_{-10}^{28}f(x)dx=(95)-(-41)-(-26)-(-38)-(62)\\&\int_{-10}^{28}f(x)dx=138\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-3}^{45}f(x)dx\\&\int_{-3}^{45}f(x)dx=\int_{-3}^{9}f(x)dx+\int_{9}^{22}f(x)dx-\int_{37}^{22}f(x)dx-\int_{39}^{37}f(x)dx-\int_{42}^{39}f(x)dx+\int_{42}^{45}f(x)dx\\&\int_{-3}^{45}f(x)dx=(-67)+(74)-(-18)-(-67)-(68)+(86)\\&\int_{-3}^{45}f(x)dx=110\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-9}^{18}f(x)dx\\&\int_{-9}^{18}f(x)dx=\int_{-9}^{0}f(x)dx-\int_{13}^{0}f(x)dx-\int_{18}^{13}f(x)dx\\&\int_{-9}^{18}f(x)dx=(62)-(-46)-(20)\\&\int_{-9}^{18}f(x)dx=88\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-10}^{56}f(x)dx\\&\int_{-10}^{56}f(x)dx=\int_{-10}^{3}f(x)dx+\int_{3}^{13}f(x)dx-\int_{23}^{13}f(x)dx+\int_{23}^{37}f(x)dx+\int_{37}^{50}f(x)dx+\int_{50}^{56}f(x)dx\\&\int_{-10}^{56}f(x)dx=(95)+(63)-(78)+(-54)+(-56)+(77)\\&\int_{-10}^{56}f(x)dx=47\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-4}^{24}f(x)dx\\&\int_{-4}^{24}f(x)dx=\int_{-4}^{6}f(x)dx-\int_{12}^{6}f(x)dx-\int_{15}^{12}f(x)dx-\int_{19}^{15}f(x)dx-\int_{22}^{19}f(x)dx-\int_{24}^{22}f(x)dx\\&\int_{-4}^{24}f(x)dx=(5)-(90)-(73)-(-70)-(97)-(36)\\&\int_{-4}^{24}f(x)dx=-221\\&\end{align*}$

$\displaystyle \begin{align*}&\text{One of the principles of integrals is one of summation. If the}\\&\text{values of an integral across certain sets of points is known,}\\&\text{this could potentially be used to find integral values at other}\\&\text{points. Imagine a set of points that are in ascending numerical}\\&\text{value: a, b, c, and d. It follows that:}\\&\int_a^d=\int_a^b+\int_b^c+\int_c^d\\&\text{Another principle of integrals is that if an integral value}\\&\text{is known, to take the same integral backwards will give a negative}\\&\text{of the original value:}\\&\int_a^b=-\int_b^a\\&\text{Knowing this, we can calculate }\int_{-4}^{32}f(x)dx\\&\int_{-4}^{32}f(x)dx=\int_{-4}^{0}f(x)dx-\int_{15}^{0}f(x)dx+\int_{15}^{23}f(x)dx+\int_{23}^{31}f(x)dx+\int_{31}^{32}f(x)dx\\&\int_{-4}^{32}f(x)dx=(-13)-(56)+(-53)+(96)+(-44)\\&\int_{-4}^{32}f(x)dx=-70\\&\end{align*}$