What is the Integral of tan(π4)ydy?

2 Answers
mason m
Oct 21, 2016

Assuming no errors in the question:

I=∫tan(π4)ydy

Since tan(π4)=1, the integral simplifies to:

I=∫ydy

I=∫y1dy

Which can be integrated using the rule: ∫yndy=yn+1n+1+C

I=y1+11+1+C

I=y22+C

mason m
Oct 21, 2016

Assuming this specific error in the question:

I=∫tan(π4y)dy

We will use substitution. First let s=π4y. Differentiating both sides yields ds=π4dy. Thus, dy=4πds.

I=∫tan(s)(4πds)

I=4π∫tan(s)ds

You may already know how to integrate tangent, but this is a reminder. Rewrite tangent using sine and cosine:

I=4π∫sin(s)cos(s)ds

Now, let t=cos(s), implying that dt=−sin(s)ds. In this case, we see that sin(s)ds=−dt.

I=4π∫−dtt

I=−4π∫dtt

This is an important integral: ∫dtt=ln|t|+C. Thus

I=−4πln|t|+C

Since t=cos(s):

I=−4πln|cos(s)|+C

Since s=π4y:

I=−4πln∣∣cos(π4y)∣∣+C

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