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Answers
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Answer:
#Be Brainly.
Explanation:
Hey there!
Answer:
\begin{gathered}\int \frac{1}{1 - tanx} .dx\\ \\= \frac{1}{1 - \frac{sinx}{cosx}} \\ \\ = \frac{1}{\frac{cosx - sinx}{cosx}} \\ \\ = \frac{cosx}{cosx - sinx} \\ \\ Multiply\: and\: divide \:by \:2\\ \\= \frac{2cosx}{2(cosx - sinx)} \\ \\ = \frac{cosx + cosx}{2(cosx - sinx)}\\ \\ Now\: adding\: and\: subtracting \:sinx \:to \:num:\\ \\= \frac{cosx + cosx + sinx - sinx}{2(cosx - sinx)} \\ \\ = \frac{(cosx + sinx) + (-cosx - sinx)}{2(cosx - sinx)}\\ \\ = \frac{1}{2}.\frac{(cosx + sinx)}{(cosx - sinx)} + \frac{1}{2}.\frac{(cosx - sinx)}{(cosx - sinx)}\\ \\ = \frac{1}{2}. (\frac{cosx + sinx}{cosx - sinx}) + \frac{1}{2}\\ \\ Now \:put \:cosx\: - sin\:x \:= \:t\\ \\ (-sinx - cosx)dx = dt\\ \\ -(cosx + sinx)dx = dt\\ \\ (cosx + sinx)dx = -dt\\ \\ Now,\: Integration\\ \\ = -\frac{1}{2} \int \frac{dt}{t} + \frac{1}{2} \int 1.dx\\ \\ = -\frac{1}{2} log |t| + \frac{x}{2} \\ \\ = -\frac{1}{2} log | cosx - sinx | + \frac{1}{2} x + C\end{gathered}
∫
1−tanx
1
.dx
=
1−
cosx
sinx
1
=
cosx
cosx−sinx
1
=
cosx−sinx
cosx
Multiplyanddivideby2
=
2(cosx−sinx)
2cosx
=
2(cosx−sinx)
cosx+cosx
Nowaddingandsubtractingsinxtonum:
=
2(cosx−sinx)
cosx+cosx+sinx−sinx
=
2(cosx−sinx)
(cosx+sinx)+(−cosx−sinx)
=
2
1
.
(cosx−sinx)
(cosx+sinx)
+
2
1
.
(cosx−sinx)
(cosx−sinx)
=
2
1
.(
cosx−sinx
cosx+sinx
)+
2
1
Nowputcosx−sinx=t
(−sinx−cosx)dx=dt
−(cosx+sinx)dx=dt
(cosx+sinx)dx=−dt
Now,Integration
=−
2
1
∫
t
dt
+
2
1
∫1.dx
=−
2
1
log∣t∣+
2
x
=−
2
1
log∣cosx−sinx∣+
2
1
x+C
#Be Brainly.