Question

Evaluate the following indefinite integrals. Show your work please!

a) ∫ tan^3 4x dx

b) ∫ sec^4 3x dx

Answer 1

$Let \ I = \int tan^{3}(4x)\ dx \\ \\= \int tan^{2}(4x) tan(4x)\ dx \\ \\= \int( sec^{2}(4x)-1)tan(4x)\ dx \\ \\ = \int sec^{2}(4x) tan(4x) \ dx-\int tan(4x) dx---(1) \\ \\Put\ tan(4x) = t\ in\ Ist\ integral \ of\ equation\ (1) \\ \\=> sec^{2}(4x).4.dx= dt \\ \\=> sec^{2}(4x) dx= \frac{dt}{4} \\ \\ put\ these\ in\ equation\ (1) \\ \\so\ I = \int t\frac{dt}{4}- \int tan(4x)dx \\ \\= \frac{1}{4}\int t\ dt-\int tan(4x) \ dx \\ \\= \frac{1}{4}(\frac{t^{2}}{2})-(-\frac{ln(cos(4x))}{4})+C \\ \\ ( since\int t^{n}dt= \frac{t^{n+1}}{n+1}+C \\ and\int tan(ax) dx= \frac{-ln(cos(ax))}{a}+ C) \\ \\ => I = \frac{t^{2}}{8}+\frac{1}{4}ln( cos(4x))+C \\ \\ Put\ t= tan(4x) \\ \\ => I = \frac{tan^{2}(4x)}{8}+\frac{1}{4}ln( cos(4x))+C \\ \\ where\ C\ is\ constant\ of\ integration\ \\ Answer.$

b)

$Let \ I = \int sec^{4}(3x)\ dx \\ \\= \int sec^{2}(3x) sec^{2}(3x)\ dx \\ \\= \int( tan^{2}(3x)+1)sec^{2}(3x)\ dx \\ \\= \int sec^{2}(3x) tan^{2}(3x) \ dx+\int sec^{2}(3x) dx---(1) \\ \\Put\ tan(3x) = t\ in\ Ist\ integral \ of\ equation\ (1) \\ \\=> sec^{2}(3x).3.dx= dt \\ \\=> sec^{2}(3x) dx= \frac{dt}{3} \\ \\ put\ these\ in\ equation\ (1) \\ \\ so\ I = \int t^{2}\frac{dt}{3}+\int sec^{2}(3x)dx \\ \\ = \frac{1}{3}\int t^{2}\ dt+\int sec^{2}(3x) \ dx \\ \\ = \frac{1}{3}(\frac{t^{3}}{3})+\frac{tan(3x)}{3}+C \\ ( since\int t^{n}dt= \frac{t^{n+1}}{n+1}+C \\ and\int sec^{2}(ax) dx= \frac{tan(ax)}{a}+ C) \\ \\ => I = \frac{t^{3}}{9}+\frac{1}{3}tan(3x)+C \\ \\ Put\ t= tan(3x) \\ \\ => I = \frac{tan^{3}(3x)}{9}+\frac{1}{3}tan(3x)+C \\ \\ where\ C\ is\ constant\ of\ integration\ \\ Answer.$