\int \tan ^5(7x).sec ^4(7x)dx
We know,
\tan ^5(7x)=\tan ^4(7x).tan(7x) i.e. applying the algebric property x^(a) = x^(a-1) . x
=\int \tan ^4(7x)\tan (7x)\sec ^4(7x)dx
Also,
\tan ^4(7x)=(\tan ^2(7x))^2
=\int(\tan ^2(7x))^2\tan(7x)\sec ^4(7x)dx
Now,using the identity : \tan ^2(x)=-1+\sec ^2(x)
=\int(-1+\sec ^2(7x))^2\tan (7x)\sec ^4(7x)dx
Applying integral substitution as: \int f(g(x))\cdot g^'(x)dx=\int f(u)du,quad u=g(x)
\sec (7x)=u\quad dx=\frac{1}{7\tan (7x)\sec(7x)}du as shown below,
Substituting sec(7x)=u
=\int (-1+u^2)^2\tan(7x)u^4\frac{1}{7\tan(7x)u}du
=\int \frac{1}{7}u^3(u^2-1)^2du
Taking the constant out as: \int a\cdot f(x)dx=a\cdot \int f(x)dx
=\frac{1}{7}\int u^3(u^2-1)^2du
Expanding u^3(u^2-1)^2
=\frac{1}{7}\int (u^7-2u^5+u^3)du
Applying sum rule as: \int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx
=\frac{1}{7}(\int u^7du-\int 2u^5du+\int u^3du)
Now,we know
\int u^7du=\frac{u^8}{8}
\int 2u^5du=\frac{u^6}{3} (2(u^6)/6 =u^6/3)
\int u^3du=\frac{u^4}{4}
i.e
=\frac{1}{7}(\frac{u^8}{8}-\frac{u^6}{3}+\frac{u^4}{4})
Substituting back u=sec(7x),
=\frac{1}{7}(\frac{\sec ^8(7x)}{8}-\frac{\sec ^6(7x)}{3}+\frac{\sec ^4(7x)}{4})
Simplifying it,
=\frac{1}{7}(\frac{1}{8}\sec ^8(7x)-\frac{1}{3}\sec ^6\left(7x)+\frac{1}{4}\sec ^4(7x))
Finally adding constant to the solution,
=\frac{1}{7}(\frac{1}{8}\sec ^8(7x)-\frac{1}{3}\sec ^6\left(7x)+\frac{1}{4}\sec ^4(7x)) + C
