2 votes 2 votes $\int_{0}^{\frac{\pi}{4}}( \sec 2x -\tan 2x )\ dx$ Calculus calculus + – PEKKA asked Jan 4, 2017 • edited Jan 4, 2017 by srestha PEKKA 611 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Samujjal Das commented Jan 4, 2017 reply Follow Share log2. 0 votes 0 votes Pavan Kumar Munnam commented Jan 4, 2017 i edited by Pavan Kumar Munnam Jan 4, 2017 reply Follow Share ----- 0 votes 0 votes PEKKA commented Jan 4, 2017 reply Follow Share please explain with steps. 0 votes 0 votes PEKKA commented Jan 4, 2017 reply Follow Share Ans given as (1/2) ln 2 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes $\int$sec2x = 1/2 * ln|sec2x + tan2x| $\int$tan2x = -1/2 * ln|cos2x| $\int_{0}^{\prod/4}$(sec2x - tan2x) = $\int_{0}^{\prod/4}$(1/2ln|sec2x + tan2x|+1/2ln|cos2x|) = $\int_{0}^{\prod/4}$(1/2ln|1+sin2x|) = 1/2ln2 Samujjal Das answered Jan 4, 2017 • edited Jan 4, 2017 by Samujjal Das Samujjal Das comment Share Follow See all 6 Comments See all 6 6 Comments reply Samujjal Das commented Jan 4, 2017 reply Follow Share please check it if there is any mistake 0 votes 0 votes PEKKA commented Jan 4, 2017 reply Follow Share answer given as (1/2) ln 2 0 votes 0 votes Pavan Kumar Munnam commented Jan 4, 2017 reply Follow Share ∫sec2x = 1/2 * log|sec2x + tan2x| ∫tan2x = - 1/2 * log|cos2x| these are correct 3 votes 3 votes Dulqar commented Jan 4, 2017 reply Follow Share @pavan is correct . @gate u did make a mistake :) it willbe 2 ln 2 0 votes 0 votes PEKKA commented Jan 4, 2017 reply Follow Share thanks :) 0 votes 0 votes Samujjal Das commented Jan 4, 2017 reply Follow Share yeah corrected!! thanks 0 votes 0 votes Please log in or register to add a comment.