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Question: How do you integrate \(\int{\dfrac{dx}{{{x}^{2}}+25}}\) using substitution?...

How do you integrate ∫dxx2+25\int{\dfrac{dx}{{{x}^{2}}+25}} using substitution?

Explanation

Solution

We can solve problems of integration like this recalling the Pythagorean identities in trigonometry. First, we start solving the problem assuming the variable xx to be 5tan⁡θ5\tan \theta and rewrite the given expression substituting the assumed value of xx . Then, we perform integration of the expression which will give us the integrated result in terms of θ\theta . Again, we substitute the value of θ\theta in the expression to get the result in terms of xx .

Complete step by step answer:
The given expression is
∫dxx2+25\int{\dfrac{dx}{{{x}^{2}}+25}}
To solve the integration by substituting trigonometric variables we must recall the Pythagorean identities in trigonometry.
1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta could be useful if we see a−bu2a-b{{u}^{2}} (We try to get k(1−sin⁡2θ)=kcos⁡2θk(1-{{\sin }^{2}}\theta )=k{{\cos }^{2}}\theta )
In this case, we have a square added to a number.
Hence, we think about using the relation between tan⁡θ\tan \theta and sec⁡θ\sec \theta as
1+tan⁡2θ=sec⁡2θ1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta which gives us our substitution.
We want to have 25tan⁡2θ+2525{{\tan }^{2}}\theta +25 , so we will use x=5tan⁡θx=5\tan \theta
Substituting the value of xx the denominator of the given expression becomes
=(5tan⁡θ)2+25={{\left( 5\tan \theta \right)}^{2}}+25
=25tan⁡2θ+25=25{{\tan }^{2}}\theta +25
Taking 2525 out common we get
=25(tan⁡2θ+1)=25\left( {{\tan }^{2}}\theta +1 \right)
=25sec⁡2θ=25{{\sec }^{2}}\theta
Also, to substitute the term dxdx we differentiate x=5tan⁡θx=5\tan \theta as shown below
⇒dx=d(5tan⁡θ)\Rightarrow dx=d\left( 5\tan \theta \right)
⇒dx=5sec⁡2θ⋅dθ\Rightarrow dx=5{{\sec }^{2}}\theta \cdot d\theta
Also, from x=5tan⁡θx=5\tan \theta we get
θ=tan⁡−1(x5)\theta ={{\tan }^{-1}}\left( \dfrac{x}{5} \right)
Therefore, the integral becomes
=∫5sec⁡2θdθ25(tan⁡2θ+1)=\int{\dfrac{5{{\sec }^{2}}\theta d\theta }{25\left( {{\tan }^{2}}\theta +1 \right)}}
=∫5sec⁡2θdθ25(sec⁡2θ)=\int{\dfrac{5{{\sec }^{2}}\theta d\theta }{25\left( {{\sec }^{2}}\theta \right)}}
Taking out the constant terms we get
=525∫dθ=\dfrac{5}{25}\int{d\theta }
=15∫dθ=\dfrac{1}{5}\int{d\theta }
Doing the integration, we get
=15θ+c=\dfrac{1}{5}\theta +c
Here, cc is a constant.
Now, we substitute the value of θ\theta in the above expression and get
=15tan⁡−1(x5)+c=\dfrac{1}{5}{{\tan }^{-1}}\left( \dfrac{x}{5} \right)+c
Therefore, by integrating the given expression using trig substitution we get 15tan⁡−1(x5)+c\dfrac{1}{5}{{\tan }^{-1}}\left( \dfrac{x}{5} \right)+c .

Note:
While doing substitution using trigonometric relation, we must choose a proper trigonometric relation for substitution. Otherwise, we will not be getting the correct result. Also, while integrating using substitution we must also substitute the dxdx term as it often gets ignored by students making the problem unsolvable.