While changing limits on integral, when using substitution , both upper and lower limits comes same, does this mean the value of integral is zero?

Question

When i did that substitution both upper and lower bound comes out to be same, i have solved the integral using trigonometry and ans is not zero.

Answer 1

The formula $$\int_a^b f(g(x))\,dx=\int_{g(a)}^{g(b)}f(y)\frac1{g'(g^{-1}(y))}\,dy$$

holds under the assumption $g:(a,b)\to\Bbb R$ is differentiable and with non-zero derivative. This is not the case for $g(\theta)=\sin(2\theta)-2$ on $(0,2\pi)$. For your substitution(s) to work you have to split the integral as $$\int_0^{2\pi}=\int_0^{\pi/4}+\int_{\pi/4}^{3\pi/4}+\int_{3\pi/4}^{5\pi/4}+\int_{5\pi/4}^{7\pi/4}+\int_{7\pi/4}^{2\pi}$$

You can also, thanks to periodicity, do $$\int_0^{2\pi}=\int_{-\pi/4}^{7\pi/4}=\int_{-\pi/4}^{\pi/4}+\int_{\pi/4}^{3\pi/4}+\int_{3\pi/4}^{5\pi/4}+\int_{5\pi/4}^{7\pi/4}$$