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I have an integral as follows:

$$\int_0^T \cos\theta\cdot dt = xT$$

where $\theta$ is a function of $t$

I also have,

$$\int_0^T \sin\theta\cdot dt = y$$

I want to solve for $T$.

If the integrals were indefinite, I could have differentiated the second one and gotten $\theta$ as a function of $t$ in terms of $y$. Integrating that would have got me $xT$, but unfortunately, I don't see how I can do that as the integrals are under limits.

I also thought of substituting $\cos\theta$ as $\sqrt{1 - \sin^2\theta}$ and then somehow simplifying so that I get it in terms of the left hand side of the second equation but was unsuccessful.

I am certain it is possible since the second integral is a sine of a function that is common to both integrals, there must be enough information contained in the second equation to completely determine the first integral. I just don't know how to extract it.

How do I go about it? I just learnt integration recently, and I'm not all that familiar with the various techniques.

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  • $\begingroup$ Do you mean $\int_0^T \dots d\,\theta?$ Or is there a function $\theta(t)$? $\endgroup$ – gammatester Jun 23 '14 at 12:59
  • $\begingroup$ So, you have 2 integrals given, and you have to solve for the upper limit $T$ . Right? $\endgroup$ – Kushashwa Ravi Shrimali Jun 23 '14 at 12:59
  • $\begingroup$ @gammatester: $\theta$ is a function of $t$. $\endgroup$ – Gerard Jun 23 '14 at 13:06
  • $\begingroup$ @KushashwaRaviShrimali: Yes. $\endgroup$ – Gerard Jun 23 '14 at 13:07
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    $\begingroup$ @enthdegree: I actually need this solution for a physics problem. I can tell you that $\theta$ varies continuously from $\pi/2$ at $t=0$ to $0$ at $t=T$. Does that help? $\endgroup$ – Gerard Jun 24 '14 at 3:04

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