5.3 Solved examples for indefinite integrals

  Basic Mathematics and Measurements

Question: Evaluate $\int {\left( {{x^4} + \frac{4}{{{x^2}}} - \frac{1}{x} - \frac{3}{{\sqrt x }} + 10} \right)} dx$?

Solution: Given $$I = \int {\left( {{x^4} + \frac{4}{{{x^2}}} - \frac{1}{x} - \frac{3}{{\sqrt x }} + 10} \right)} dx$$$$I = \int {\left( {{x^4} + 4{x^{ - 2}} - \frac{1}{x} - 3{x^{ - \frac{1}{2}}} + 10} \right)} dx$$$$I = \frac{{{x^5}}}{5} - \frac{4}{x} - \ln x - \frac{3}{{\frac{3}{2}}}{x^{ - \frac{3}{2}}} + 10x + c$$$$I = \frac{{{x^5}}}{5} - \frac{4}{x} - \ln x - \frac{2}{{{x^{\frac{3}{2}}}}} + 10x + c$$

Question: Evaluate $\int {x\sin xdx} $?

Solution: The above integration is a multiplication of two functions.

We will use integrations by parts for this type of integrations.

$$\int {uvdx} = u\int {vdx} - \int {\frac{{du}}{{dx}}\left( {\int {vdx} } \right)} \,dx$$
So, $$\int {x\sin xdx} = x\int {\sin xdx} - \int {\frac{{dx}}{{dx}}} \left( {\int {\sin xdx} } \right)dx$$$$\int {x\sin xdx} = - x\cos x - \int {1.} \left( { - \cos x} \right)dx$$$$\int {x\sin xdx} = - x\cos x + \int {\cos xdx} $$$$\int {x\sin xdx} = - x\cos x + \sin x + c$$

Question: Evaluate $\int {{{\sin }^3}x} dx$ ?

Solution: $$I = \int {{{\sin }^3}x} dx$$$$I = \int {\sin x\left( {{{\sin }^2}x} \right)} dx$$$$I = \int {\sin x\left( {1 - {{\cos }^2}x} \right)} dx$$Let $$\cos x = u$$$$ - \sin x = \frac{{du}}{{dx}}$$$$ - du = \sin xdx$$Therefore, $$I = - \int {\left( {1 - {u^2}} \right)du} $$$$I = - u + \frac{{{u^3}}}{3} + c$$$$I = - \cos x + \frac{{{{\cos }^3}x}}{3} + c$$

Question: Evaluate $\int {\frac{{dx}}{{\sqrt {3x + 4} }}} $?

Solution: $$I = \int {\frac{{dx}}{{\sqrt {3x + 4} }}} $$Let, $$3x + 4 = t$$$$3dx = dt$$$$dx = \frac{{dt}}{3}$$$$I = \frac{1}{3}\int {\frac{{dt}}{{\sqrt t }}} $$$$I = \frac{1}{3}\int {{t^{ - \frac{1}{2}}}dt} $$$$I = \frac{1}{3}\left( {\frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}}} \right) + c$$$$I = \frac{2}{3}\sqrt t + c$$$$I = \frac{2}{3}\sqrt {3x + 4} + c$$