Question

please solve this problem​

Answer 1

Answer:

1 =dx*cosx-cos²x

Step-by-step explanation:

i hope it vill help u

Answer 2

Answer:

I = $\int{\frac{1}{cosx-cos^{2}x}}dx$ = $ln|secx+tanx|-cotx-cosecx+c$

Step-by-step explanation:

Let I = $\int{\frac{1}{cosx-cos^{2}x}}dx$

or I = $\int{\frac{1}{cosx(1-cosx)}}dx$

or I = $\int{[\frac{1}{cosx}+\frac{1}{(1-cosx)}]}dx$

or I = $\int{[secx+\frac{1.(1+cosx)}{(1-cosx)(1+cosx)}]}dx$

or I = $\int{[secx+\frac{1+cosx}{1-cos^{2}x}]}dx$

or I = $\int{[secx+\frac{1+cosx}{sin^{2}x}]}dx$

or I = $\int{[secx+\frac{1}{sin^{2}x}+\frac{cosx}{sin^{2}x}]}dx$

or I = $\int{[secx+cosec^{2}x+cosecx.cotx]}dx$

or I = $\int{secx}dx+\int{cosec^{2}x}dx+\int{cosecx.cotx}dx$

or I = $ln|secx+tanx|-cotx-cosecx+c$

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