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paulmdrdo1
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what method should i use here(except integration by parts)?
\begin{align*}\displaystyle \int sec^3\theta \,d\theta\end{align*}
\begin{align*}\displaystyle \int sec^3\theta \,d\theta\end{align*}
paulmdrdo said:what method should i use here(except integration by parts)?
\begin{align*}\displaystyle \int sec^3\theta \,d\theta\end{align*}
To integrate sec^3x, you can use the substitution method. Let u = tanx, then du = sec^2x dx. Substituting these values into the integral, you get ∫sec^3x dx = ∫sec^3x * sec^2x dx = ∫(u^2)(du) = (u^3)/3 + C = (tan^3x)/3 + C.
The general formula for integrating sec^3x is ∫sec^3x dx = (secx)(tanx) + ln|secx + tanx| + C.
Yes, you can use integration by parts to integrate sec^3x. Let u = secx and dv = sec^2x dx, then du = secx tanx dx and v = tanx. Substituting these values into the integration by parts formula, you get ∫sec^3x dx = secx * tanx - ∫secx tan^2x dx. Using the identity tan^2x = sec^2x - 1, the integral becomes ∫sec^3x dx = secx * tanx - ∫secx (sec^2x - 1) dx = secx * tanx - ∫sec^3x dx + ∫secx dx. Solving for the original integral, you get ∫sec^3x dx = (secx)(tanx) + ln|secx + tanx| + C.
Yes, you can use the substitution tanx = u to integrate sec^3x. This will transform the integral into ∫sec^3x dx = ∫(1 + tan^2x) secx dx = ∫(1 + u^2) du = u + (u^3)/3 + C = tanx + (tan^3x)/3 + C.
Yes, the integral of sec^3x can be solved using partial fractions. By performing partial fraction decomposition, you can rewrite the integral as ∫sec^3x dx = ∫(A/(secx + 1)) + (B/(secx - 1)) dx. Solving for A and B and performing the integrals, you get ∫sec^3x dx = (ln|secx + 1| - ln|secx - 1|) + C = ln|secx + tanx| + C.