Answer to Question #101489 in Calculus for Mal

Question

1. Integrate with the placement method: ∫cosxsin2xdx

2. Integrate using the partial integration method
a.) ∫e^x*cosx dx
b.) ∫x^2*e^3x dx

Answer 1

Solution.

1. Using the formula

$$sin2x=2sinxcosx$$

get

$$\intop cosxsin2xdx=\intop 2(cosx)^2sinxdx$$

We use a replacement k=cosx, therefore dk=-sinxdx. As result making a replacement, we get

$$\intop 2(cosx)^2sinxdx=-\intop 2k^2dk=-\frac {2k^3}{3}+C$$

where C is constant. Returning to the replacement, we get

$$\intop cosxsin2xdx=-\frac{2(cosx)^3}{3} +C$$

2. Integrate using the partial integration method

a)

$$\begin{vmatrix} u=e^x & dv=cosxdx \\ du=e^xdx & v=sinx \end{vmatrix}$$

Therefore we write the integral as

$$\intop e^xcosxdx=e^xsinx-\intop e^xsinxdx$$

Using the partial integration method get

$$\begin{vmatrix} u=e^x & dv=sinxdx \\ du=e^xdx & v=-cosx \end{vmatrix}$$

$$e^xsinx-\intop e^xsinxdx=e^xsinx+e^xcosx-\intop e^xcosx$$

Hence

$$\intop e^xcosxdx=\frac{e^xsinx+e^xcosx}{2}+C$$

where C is constant.

b. Using the partial integration method get

$$\begin{vmatrix} u=x^2 & dv=e^{3x}dx \\ du=2xdx & v=\frac{e^{3x}}{3} \end{vmatrix}$$

Therefore we write the integral as

$$\intop x^2e^{3x}dx=\frac{x^2e^{3x}}{3} -\frac{2}{3}\intop xe^{3x}dx$$

Using the partial integration method get

$$\begin{vmatrix} u=x & dv=e^{3x}dx \\ du=dx & v=\frac{e^{3x}}{3} \end{vmatrix}$$

Therefore we write the integral as

$$\intop x^2e^{3x}dx=\frac{x^2e^{3x}}{3} -\frac{2xe^{3x}}{9} +\frac{2}{9}\intop e^{3x}dx$$

As result

$$\intop x^2e^{3x}dx=\frac{x^2e^{3x}}{3} -\frac{2xe^{3x}}{9} +\frac{2e^{3x}}{27}+C$$

where C is constant.

Answer. 1.

$$\intop cosxsin2xdx=-\frac{2(cosx)^3}{3} +C$$

where C is constant.

2 a.

$$\intop e^xcosxdx=\frac{e^xsinx+e^xcosx}{2}+C$$

where C is constant.

2 b.

$$\intop x^2e^{3x}dx=\frac{x^2e^{3x}}{3} -\frac{2xe^{3x}}{9} +\frac{2e^{3x}}{27}+C$$

where C is constant.