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What is the Integral of # (1+tanx)^5 * (sec^2(x)) dx #?

Answer 1

Paisley Johnson

The answer is #=(1+tanx)^6/6+C#

We must calculate #int(1+tanx)^5sec^2xdx#

We do this integral by substitution

Let #u=1+tanx#

#du=sec^2xdx#

#int(1+tanx)^5sec^2xdx#

#=intu^5du=u^6/6#

#=(1+tanx)^6/6+C#

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Answer 2

Evelyn Agard

To integrate the given expression ( \int (1 + \tan(x))^5 \cdot \sec^2(x) , dx ), use the substitution method. Let ( u = 1 + \tan(x) ), then ( du = \sec^2(x) , dx ). Substituting these into the integral, you get ( \int u^5 , du ). Now integrate ( u^5 ) with respect to ( u ) to get ( \frac{u^6}{6} + C ), where ( C ) is the constant of integration. Finally, substitute back ( u = 1 + \tan(x) ) to get the final result ( \frac{(1 + \tan(x))^6}{6} + C ).

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Answer 3

Paisley Johnson

The integral of (1 + tan(x))^5 * (sec^2(x)) dx equals (1/6) * (1 + tan(x))^6 + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.