Question

resuelve, por favor es álgebra​

Answer 1

Respuesta:

C

Explicación paso a paso:

$\tan ^2\left(x\right)-\sin ^2\left(x\right)=\tan ^2\left(x\right)\sin ^2\left(x\right)=\left(1-\cos ^2\left(x\right)\right)\tan ^2\left(x\right)=\tan ^2\left(x\right)-\cos ^2\left(x\right)\tan ^2\left(x\right)=\tan ^2\left(x\right)-\cos ^2\left(x\right)\tan$$x^2-x-6=0x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \left(-6\right)}}{2\cdot \:1}x_{1,\:2}=\frac{-\left(-1\right)\pm \:5}{2\cdot \:1}x_1=\frac{-\left(-1\right)+5}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-5}{2\cdot \:1}x=3,\:x=-2\int \:e^x\cos \left(x\right)dx$$=e^x\sin \left(x\right)-\left(-e^x\cos \left(x\right)-\int \:-e^x\cos \left(x\right)dx\right)=e^x\sin \left(x\right)-\left(-e^x\cos \left(x\right)-\int \:-e^x\cos \left(x\right)dx\right)=\frac{e^x\sin \left(x\right)}{2}+\frac{e^x\cos \left(x\right)}{2}=\frac{e^x\sin \left(x\right)}{2}+\frac{e^x\cos \left(x\right)}{2}+C$