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  • https://ukrayinska.libretexts.org/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0/%D1%80%D0%BE%D0%B7%D1%80%D0%B0%D1%85%D1%83%D0%BD%D0%BA%D1%83/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%9E%D0%B1%D1%87%D0%B8%D1%81%D0%BB%D0%B5%D0%BD%D0%BD%D1%8F_(OpenStax)/18%3A_%D0%94%D0%BE%D0%B4%D0%B0%D1%82%D0%BA%D0%B8/18.B%3A_%D0%A2%D0%B0%D0%B1%D0%BB%D0%B8%D1%86%D1%8F_%D1%96%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%96%D0%B2
    39. \(\quad \displaystyle ∫u^n\sin u\,du=−u^n\cos u+n∫u^{n−1}\cos u\,du\) 40. \(\quad \displaystyle ∫u^n\cos u\,du=u^n\sin u−n∫u^{n−1}\sin u\,du\) 41. \(\quad \begin{align*} \displaystyle ∫\sin^n u\co...39. \(\quad \displaystyle ∫u^n\sin u\,du=−u^n\cos u+n∫u^{n−1}\cos u\,du\) 40. \(\quad \displaystyle ∫u^n\cos u\,du=u^n\sin u−n∫u^{n−1}\sin u\,du\) 41. \(\quad \begin{align*} \displaystyle ∫\sin^n u\cos^m u\,du = −\frac{\sin^{n−1}u\cos^{m+1}u}{n+m}+\frac{n−1}{n+m}∫\sin^{n−2}u\cos^m u\,du \\[4pt] =\frac{\sin^{n+1}u\cos^{m−1}u}{n+m}+\frac{m−1}{n+m}∫\sin^n u\cos^{m−2}u \,du \end{align*}\)