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Question: How do you find the integral of \(\int{\sin x\tan x dx}\)?...

How do you find the integral of ∫sin⁡xtan⁡xdx\int{\sin x\tan x dx}?

Explanation

Solution

We can not directly integrate the given trigonometric function. We first try to simplify the function where we express the multiplication form in the form addition and subtraction. We then integrate the new form of trigonometric function. We use ∫sec⁡xdx=log⁡∣sec⁡x+tan⁡x∣+c\int{\sec x dx}=\log \left| \sec x+\tan x \right|+c and ∫cos⁡xdx=sin⁡x+c\int{\cos xdx}=\sin x+c.

Complete step-by-step solution:
First, we find the simplified form of the given trigonometric function sin⁡xtan⁡x\sin x\tan x. Then we try to find the integration of the simplified form.
We have the identity theorem for ratio tan where tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
We multiply sin⁡x\sin x to the both sides of the equation tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}.
⇒sin⁡xtan⁡x=sin⁡xcos⁡x×sin⁡x=sin⁡2xcos⁡x\Rightarrow \sin x\tan x=\dfrac{\sin x}{\cos x}\times \sin x=\dfrac{{{\sin }^{2}}x}{\cos x}.
We know that the sum of squares of ratio sin and cos is always 1.
This gives sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. We replace sin⁡2x{{\sin }^{2}}x with sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x.
⇒sin⁡xtan⁡x=1−cos⁡2xcos⁡x\Rightarrow \sin x\tan x=\dfrac{1-{{\cos }^{2}}x}{\cos x}.
Breaking the division, we get sin⁡xtan⁡x=1cos⁡x−cos⁡2xcos⁡x=sec⁡x−cos⁡x\sin x\tan x=\dfrac{1}{\cos x}-\dfrac{{{\cos }^{2}}x}{\cos x}=\sec x-\cos x.
Now we find the simplified form of the trigonometric ratio where sin⁡xtan⁡x=sec⁡x−cos⁡x\sin x\tan x=\sec x-\cos x.
Taking integration on both sides, we get ∫sin⁡xtan⁡xdx=∫(sec⁡x−cos⁡x)dx\int{\sin x\tan x dx}=\int{\left( \sec x-\cos x \right)dx}.
We know that ∫sec⁡xdx=log⁡∣sec⁡x+tan⁡x∣+c\int{\sec x dx}=\log \left| \sec x+\tan x \right|+c and ∫cos⁡xdx=sin⁡x+c\int{\cos xdx}=\sin x+c.
We put these values and get
∫sin⁡xtan⁡xdx=∫(sec⁡x−cos⁡x)dx =∫sec⁡xdx−∫cos⁡xdx =log⁡∣sec⁡x+tan⁡x∣−sin⁡x+c \begin{aligned} & \int{\sin x\tan x dx}=\int{\left( \sec x-\cos x \right)dx} \\\ & =\int{\sec x dx}-\int{\cos xdx} \\\ & =\log \left| \sec x+\tan x \right|-\sin x+c \\\ \end{aligned}
Here cc is the integral constant.
Therefore, the integral of ∫sin⁡xtan⁡xdx\int{\sin x\tan x dx} is log⁡∣sec⁡x+tan⁡x∣−sin⁡x+c\log \left| \sec x+\tan x \right|-\sin x+c.

Note: These integrals are evaluated by applying trigonometric identities, as outlined in the particular rules. The integration works for different ratios involving the by parts theorem also. The differential is always equal to the anti-derivative law.