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We have:
I=∫csc3xdx
We will use integration by parts. First, rewrite the integral as:
I=∫csc2xcscxdx
Since integration by parts takes the form ∫udv=uv−∫vdu, let:
{u=cscx ⇒ du=−cotxcscxdxdv=csc2xdx ⇒ v=−cotx
Applying integration by parts:
I=−cotxcscx−∫cot2xcscxdx
Through the Pythagorean identity, write cot2xas csc2x−1.
I=−cotxcscx−∫(csc2x−1)(cscx)dx
I=−cotxcscx−∫csc3xdx+∫cscxdx
Note that I=∫csc3xdx and ∫cscxdx=−ln(|cotx+cscx|).
I=−cotxcscx−I−ln(|cotx+cscx|)
Add the original integral I to both sides.
2I=−cotxcscx−ln(|cotx+cscx|)
Solve for I and add the constant of integration:
I=−cotxcscx−ln(|cotx+cscx|)2+C
.........Be always brainliest......
......... Thanks.........
We have:
I=∫csc3xdx
We will use integration by parts. First, rewrite the integral as:
I=∫csc2xcscxdx
Since integration by parts takes the form ∫udv=uv−∫vdu, let:
{u=cscx ⇒ du=−cotxcscxdxdv=csc2xdx ⇒ v=−cotx
Applying integration by parts:
I=−cotxcscx−∫cot2xcscxdx
Through the Pythagorean identity, write cot2xas csc2x−1.
I=−cotxcscx−∫(csc2x−1)(cscx)dx
I=−cotxcscx−∫csc3xdx+∫cscxdx
Note that I=∫csc3xdx and ∫cscxdx=−ln(|cotx+cscx|).
I=−cotxcscx−I−ln(|cotx+cscx|)
Add the original integral I to both sides.
2I=−cotxcscx−ln(|cotx+cscx|)
Solve for I and add the constant of integration:
I=−cotxcscx−ln(|cotx+cscx|)2+C
.........Be always brainliest......
......... Thanks.........
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