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Question: Solve the given integral \( \int {\dfrac{{{{\sec }^2}\theta }}{{\tan \theta }}d\theta } \)...

Solve the given integral ∫sec⁡2θtan⁡θdθ\int {\dfrac{{{{\sec }^2}\theta }}{{\tan \theta }}d\theta }

Explanation

Solution

Hint : This question is of integration. Integration has many formulas and many methods to solve the problem. Given a problem contains trigonometric ratios we can go about solving the problem by reducing the problem into simple form.

Complete step-by-step answer :
Given,
∫sec⁡2θtan⁡θdθ\int {\dfrac{{{{\sec }^2}\theta }}{{\tan \theta }}d\theta }
As we know,
sec⁡θ=1cos⁡θ tan⁡θ=sin⁡θcos⁡θ   \sec \theta = \dfrac{1}{{\cos \theta }} \\\ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \;
From the given problem
=sec⁡2θtan⁡θ=cos⁡θsin⁡θ×cos⁡2θ ⇒1sin⁡θcos⁡θ    = \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} = \dfrac{{\cos \theta }}{{\sin \theta \times {{\cos }^2}\theta }} \\\ \Rightarrow \dfrac{1}{{\sin \theta \cos \theta }} \; \\\
Since,
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
We get,
=sec⁡2θtan⁡θ=cos⁡ec2θ= \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} = \cos ec2\theta
Therefore, we can write our problem as
=∫cos⁡ec2θ dθ ⇒−ln⁡∣cos⁡ec2θ+cot⁡2θ∣2+C   = \int {\cos ec2\theta \,d\theta } \\\ \Rightarrow \dfrac{{ - \ln \left| {\cos ec2\theta + \cot 2\theta } \right|}}{2} + C \;
The above is the answer to the given question.
So, the correct answer is “−ln⁡∣cos⁡ec2θ+cot⁡2θ∣2+C\dfrac{{ - \ln \left| {\cos ec2\theta + \cot 2\theta } \right|}}{2} + C ”.

Note : Students need to know the trigonometric identities and it’s conversions. It allows reduction of a given problem into the simple form and then integrates to get the solution.
Mostly they reduce to the form that has its standard integral solution.