integration root of cotx
Answers
Answered by
3
let cot x = p²
(cot x) dx = p² dp
-cosec² x dx = 2p dp
∫[(-2pdp)/(cosec²x)] = -∫[(2pdp)/(1+cot²x)
-∫[(2pdp)/(1+p⁴)
let p²=q
2pdp=dq
-∫[(dq)/(1+q²)] {∫1/(a²+x²) dx = (1/a) tan⁻¹ (x/a) + c}
-tan⁻¹(q) = -tan⁻¹(p²) = -tan⁻¹(cot x) = -tan⁻¹[tan{(π/2)-x}]
-[(π/2)-x] = x - π/2 + c
(cot x) dx = p² dp
-cosec² x dx = 2p dp
∫[(-2pdp)/(cosec²x)] = -∫[(2pdp)/(1+cot²x)
-∫[(2pdp)/(1+p⁴)
let p²=q
2pdp=dq
-∫[(dq)/(1+q²)] {∫1/(a²+x²) dx = (1/a) tan⁻¹ (x/a) + c}
-tan⁻¹(q) = -tan⁻¹(p²) = -tan⁻¹(cot x) = -tan⁻¹[tan{(π/2)-x}]
-[(π/2)-x] = x - π/2 + c
Similar questions