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Question: Evaluate the following integrals \(\int {{{\sec }^4}x\tan xdx} \)...

Evaluate the following integrals ∫sec⁡4xtan⁡xdx\int {{{\sec }^4}x\tan xdx}

Explanation

Solution

Before attempting this question, one should have prior knowledge about the concept of integration and also remember to use 1+tan⁡2x=sec⁡2x1 + {\tan ^2}x = {\sec ^2}x and take tan⁡x\tan x as t and use the formula ∫xn=xn+1n+1+C\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C, using this information you can approach the solution.

Complete step-by-step solution:
According to the given information we have integrals as ∫sec⁡4xtan⁡xdx\int {{{\sec }^4}x\tan xdx}
We can rewrite the given integrals as ∫sec⁡2xsec⁡2xtan⁡xdx\int {{{\sec }^2}x{{\sec }^2}x\tan xdx}
As we know that by the trigonometric identity 1+tan⁡2x=sec⁡2x1 + {\tan ^2}x = {\sec ^2}x
Therefore, ∫(1+tan⁡2x)sec⁡2xtan⁡xdx\int {\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x\tan xdx} (equation 1)
Let tan⁡x\tan x= t
Differentiating both side with respect to x we get
ddx(tan⁡x)=dtdx\dfrac{d}{{dx}}\left( {\tan x} \right) = \dfrac{{dt}}{{dx}}
We know that ddx(tan⁡x)=sec⁡2x\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x
So, sec⁡2x=dtdx{\sec ^2}x = \dfrac{{dt}}{{dx}}
\Rightarrow $$$$dx = \dfrac{{dt}}{{{{\sec }^2}x}}
Substituting the values in equation 1 we get
∫(1+t2)t×sec⁡2x×dtsec⁡2x\int {\left( {1 + {t^2}} \right)t \times {{\sec }^2}x \times \dfrac{{dt}}{{{{\sec }^2}x}}}
\Rightarrow $$$\int {\left( {1 + {t^2}} \right)tdt} $ \Rightarrow \int {\left( {t + {t^3}} \right)dt} $ $$ \Rightarrow \int {tdt} + \int {{t^3}dt} Weknowthatbytheintegrationidentity We know that by the integration identity\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + CTherefore, Therefore,\int {tdt} + \int {{t^3}dt} = \dfrac{{{t^{1 + 1}}}}{{1 + 1}} + \dfrac{{{t^{3 + 1}}}}{{3 + 1}} $$ \Rightarrow $$$\dfrac{{{t^2}}}{2} + \dfrac{{{t^4}}}{4} + C
Now substituting the value of t in the above equation we get
tan⁡2x2+tan⁡4x4+C\dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C
Therefore, the integral of ∫sec⁡4xtan⁡xdx=tan⁡2x2+tan⁡4x4+C\int {{{\sec }^4}x\tan xdx} = \dfrac{{{{\tan }^2}x}}{2} + \dfrac{{{{\tan }^4}x}}{4} + C.

Note: In the above question we used the method of integration to find the integral of trigonometric functions where the term “function” can be explained as a relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by f:X→Yf:X \to Y examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.