5.8-bo'lim: Aniq integrallarni almashtirish qoidasi
Iloji bo'lsa, quyidagi integrallarning har birini baholang. Agar buning iloji bo'lmasa, nima uchun integralni baholash mumkin emasligini aniq tushuntiring.
- \( \displaystyle \int_{0}^{1}{{3\left( {4x + {x^4}} \right){{\left( {10{x^2} + {x^5} - 2} \right)}^6}\,dx}}\) Yechim
- \( \displaystyle \int_{0}^{{\frac{\pi }{4}}}{{\frac{{8\cos \left( {2t} \right)}}{{\sqrt {9 - 5\sin \left( {2t} \right)} }}\,dt}}\) Yechim
- \( \displaystyle \int_{\pi }^{0}{{\sin \left( z \right){{\cos }^3}\left( z \right)\,dz}}\) Yechim
- \( \displaystyle \int_{1}^{4}{{\sqrt w \,{{\bf{e}}^{1 - \sqrt {{w^{\,3}}} }}\,dw}}\) Yechim
- \( \displaystyle \int_{{ - 4}}^{{ - 1}}{{\sqrt[3]{{5 - 2y}} + \frac{7}{{5 - 2y}}\,dy}}\) Yechim
- \( \displaystyle \int_{{ - 1}}^{2}{{{x^3} + {{\bf{e}}^{\frac{1}{4}x}}\,dx}}\) Yechim
- \( \displaystyle \int_{\pi }^{{\frac{{3\pi }}{2}}}{{6\sin \left( {2w} \right) - 7\cos \left( w \right)dw}}\) Yechim
- \( \displaystyle \int_{1}^{5}{{\frac{{2{x^3} + x}}{{{x^4} + {x^2} + 1}} - \frac{x}{{{x^2} - 4}}\,dx}}\) Yechim
- \( \displaystyle \int_{{ - 2}}^{0}{{t\sqrt {3 + {t^2}} + \frac{3}{{{{\left( {6t - 1} \right)}^2}}}\,dt}}\) Yechim
- \( \displaystyle \int_{{ - 2}}^{1}{{{{\left( {2 - z} \right)}^3} + \sin \left( {\pi z} \right){{\left[ {3 + 2\cos \left( {\pi z} \right)} \right]}^3}\,dz}}\) Yechim