#\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
