Evaluating ∫dx4−9x2 with different trig substitutions (sin vs cos) gives different resultsFor the first trig sub, I set 9x2=4cos2θ. This simplifies to: x=23cos⁡θ, and dx=−23sin⁡θ. Substituting in, I get:∫−2sin⁡θ6sin⁡θ=−θ3=−13cos−1(3x2)+CFor the second trig sub, I set 9x2=4sin2θ. This simplifies to: x=23sin⁡θ, and dx=23cos⁡θ. Substituting in, I get:∫2cos⁡θ6cos⁡θ=θ3=13sin−1(3x2)+C (2)Why do these two trig substitutions yield different results graphically? Shouldn't they result in the same graph?

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Evaluating ∫dx4−9x2 with different trig substitutions (sin vs cos) gives different resultsFor the first trig sub, I set 9x2=4cos2θ. This simplifies to: x=23cos⁡θ, and dx=−23sin⁡θ. Substituting in, I get:∫−2sin⁡θ6sin⁡θ=−θ3=−13cos−1(3x2)+CFor the second trig sub, I set 9x2=4sin2θ. This simplifies to: x=23sin⁡θ, and dx=23cos⁡θ. Substituting in, I get:∫2cos⁡θ6cos⁡θ=θ3=13sin−1(3x2)+C (2)Why do these two trig substitutions yield different results graphically? Shouldn&#039;t they result in the same graph?

Prince Huang · Accepted Answer

Since (−arcsin)′(x)=arccos′⁡(x)=−11−x2, you got twice the same thing

Evaluating \int \frac{dx}{\sqrt{4-9x^2}} with different trig substitutions (sin vs cos) gives

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