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karush
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$\int\frac{1}{\sqrt{2+3y^2}}dy$
$u=\sqrt{3/2}\tan\left({\theta}\right)$
I continued but it went south..
$u=\sqrt{3/2}\tan\left({\theta}\right)$
I continued but it went south..
Prove It said:Even though the trig sub will work, the integral will still be quite difficult. $\displaystyle \begin{align*} y = \sqrt{\frac{2}{3}}\sinh{(t)} \end{align*}$ is a much easier sub...
karush said:$\int\frac{1}{\sqrt{2+3y^2}}dy$
$u=\sqrt{3/2}\tan\left({\theta}\right)$
I continued but it went south..
karush said:$\frac{1}{\sqrt{3}}\ln\left({\sqrt{2+3y^2}+\sqrt{3}y}\right)+C$
An indefinite integral using trig substitutions is a method used to solve integrals that involve trigonometric functions. It involves substituting a trigonometric expression for a variable in the integral, which can simplify the integration process.
Trig substitutions are useful for solving integrals that involve expressions with square roots, or integrals that contain sums or differences of squares. It can also be used for integrals involving trigonometric functions raised to powers.
There are three main trig substitutions: sine, cosine, and tangent. You should choose the substitution that will result in the simplest integral. This can be determined by looking at the expression inside the integral and comparing it to the trig identities.
No, not all indefinite integrals can be solved using trig substitutions. It is most effective for integrals that involve trigonometric functions, but other methods may be needed for other types of integrals.
Yes, there are a few special cases to be aware of. For example, if the integral contains a term with a coefficient in front of the trigonometric function, this should be factored out before substituting. Also, be cautious of the signs of the trig functions, as they can affect the final result.