How Can U-Substitution Simplify Trigonometric Integrals?

jtt
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Homework Statement


use substitution to evaluate the integral


Homework Equations


1)∫ tan(4x+2)dx
2)∫3(sin x)^-2 dx

The Attempt at a Solution


1) u= 4x+2 du= 4
(1/4)∫4 tan(4x+2) dx
∫(1/4)tan(4x+2)(4dx)
∫ (1/4) tanu du
(1/4)ln ltan(u)l +c

2) u=sinx du= cosx or u=x du = 1 ?
 
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jtt said:

Homework Statement


use substitution to evaluate the integral

Homework Equations


1)∫ tan(4x+2)dx
2)∫3(sin x)^-2 dx

The Attempt at a Solution


1) u= 4x+2 du= 4
(1/4)∫4 tan(4x+2) dx
∫(1/4)tan(4x+2)(4dx)
∫ (1/4) tanu du
(1/4)ln ltan(u)l +c

2) u=sinx du= cosx or u=x du = 1 ?

Homework Statement


Homework Equations


The Attempt at a Solution


For your first integral, you evaluated ∫tan(u)du incorrectly.
∫tan(u) du = ∫sin(u)/cos(u) du
= -∫-sin(u)/cos(u) du
So now solve for this integral, given that ∫f'(x)/f(x) dx = ln(|f(x)|) + c

For your second, I'm not sure why you would use 'u' substitution,
because 1/sin^2(x) = csc^2(x), which has the integral of -cot(x) + c.

I'll leave that to you to find a way with u-substitution.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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