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Question

Question: Consider the expression\[\int{{{\sec }^{n}}x\tan xdx}\]. Find the value of the integral....

Consider the expressionsecnxtanxdx\int{{{\sec }^{n}}x\tan xdx}. Find the value of the integral.

Explanation

Solution

Hint: You can rewrite secnx{{\sec }^{n}}xas secn1x.secx{{\sec }^{n-1}}x.\sec x, in the given integral. Later, you can employ the substitution method to compute the given integral, by substitutingsecx=t\sec x=t.

We must evaluate the integral of secnxtanxdx\int{{{\sec }^{n}}x\tan xdx}.
Let us assume the given integral as secnxtanxdx=I\int{{{\sec }^{n}}x\tan xdx}=I.
We can rewrite secnx{{\sec }^{n}}xas secn1x.secx{{\sec }^{n-1}}x.\sec x, since we know am.an=am+n{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}
Therefore, the integer can be expressed as,
I=secn1x.secx.tanxdxI=\int{{{\sec }^{n-1}}x.\sec x.tanxdx}.
Let us use the substitution process for evaluating this particular form of integral.
So, let us put secx\sec xas ‘t’.
secx=t\sec x=t.
Differentiating on both the sides of the above equation, we have:
d(secx)dx=dtdx\dfrac{d\left( \sec x \right)}{dx}=\dfrac{dt}{dx}
secx.tanx=dtdx\sec x.\tan x=\dfrac{dt}{dx}
Therefore, dt=secx.tanx.dxdt=\sec x.\tan x.dx
As we substitute the value of ‘t’ and dtdtin the integrals I, the integral will transform as mentioned below:
I=(t)n1dtI=\int{{{\left( t \right)}^{n-1}}dt}
Evaluating the integral further, we have:
I=t(n1)+1(n1)+1+cI=\dfrac{{{t}^{\left( n-1 \right)+1}}}{\left( n-1 \right)+1}+c.
Since, for any given x,xndx=xn+1n+1+c\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}
Therefore, the integral reduces to:
I=tnn+cI=\dfrac{{{t}^{n}}}{n}+c
As, we have t=secxt=\sec x, let us put it back in the solved expression of the integral.
Then, we have:
I=secnxn+cI=\dfrac{{{\sec }^{n}}x}{n}+c
Where, c is any constant.
So, by following the process of substitution we have evaluated the given integral.
Hence the answer for the given integral is secnxn+c.\dfrac{{{\sec }^{n}}x}{n}+c.

Note: We can directly evaluate the given integral by using the formula (f(x))nf(x)dx=f(x)n+1n+1+c\int{{{\left( f\left( x \right) \right)}^{n}}{f}'\left( x \right)dx=\dfrac{f{{\left( x \right)}^{n+1}}}{n+1}+c} where f(x)=secn1f\left( x \right)={{\sec }^{n-1}}and f(x)=secxtanx{f}'\left( x \right)=\sec x\tan x respectively. Using shortcut methods effectively will save time and give a smart approach to the answer. Also, applying product rule is not recommended in this case as the process will be very lengthy and difficult to solve.