Question

sec² x
2-3sin x
19.
dx
20.
cosec x
12-23
dx
COS X
उत्तर प्रदेश मद्रण वर्ष 2020-21​

Answer 1

$\begin{gathered}\bf{I=\int{\frac{sec^2x}{cosec^2}}\,dx}\\\\=\bf{\int{\frac{\frac{1}{cos^2x}}{\frac{1}{sin^2}}}\,dx}\\\\=\bf{\int{\frac{sin^2x}{cos^2}}\,dx}\\\\=\bf{\int{tan^2x}\,dx}\\\\=\bf{\int{(sec^2x-1)}\,dx}\\\\=\bf{\int{sec^2x}\,dx-\int{dx}}\\\\=\bf{tanx-x+C}\end{gathered}$

• Hence, I = tanx - x + C